A solution containing a mixture of lactic acid and lactate was found to have a ph of 4.10.4.10. calculate the ratio of the lactate concentration to the lactic acid concentration in this solution.

Mass spectrometers may be used to pick out unknown compounds through molecular weight determination, to quantify regarded compounds, and to decide shape and chemical residences of molecules.

The relative abundance of every isotope may be decided the usage of mass spectrometry. A mass spectrometer ionizes atoms and molecules with a high-power electron beam after which deflects the ions via a magnetic discipline primarily based totally on their mass-to-charge ratios ( m / z m/z m/z ). In a pure sample of an element, the mass of that element is represented as an m/z ratio and can be used to identify the element. Also, identification of an element can be done by calculating the average atomic mass from the mass spectrum data.

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The latest estimate for the annual number of photos taken is 1.72 trillion.

People are taking more images all around the world, and Rise Above Research, a consulting organization that conducts market research for the digital imaging sector, predicts that the total number of photos taken worldwide will reach 1.5 trillion in 2022.

Around the world, 1.72 trillion images are taken annually, or 54,400 each second or 4.7 billion per day. Every year, around 2.3 trillion pictures will be taken by 2030. Photutorial research indicates that 1.2 trillion photos were shot globally in 2021. In 2022, the amount will rise to 1.72 trillion.

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Answer:

phosphorus trichloride

Salts and organic solvents can be used at high concentrations to dampen out electrostatic interactions among amino acid residues.

Electrostatic interactions between amino acid residues can influence the structure and function of proteins. These interactions arise from the charged nature of amino acids, where some have positively charged side chains (e.g., lysine) and others have negatively charged side chains (e.g., aspartic acid). In order to study the effect of these interactions or minimize their impact, high concentrations of certain compounds can be employed.

Salts, such as sodium chloride (NaCl), potassium chloride (KCl), or ammonium sulfate ((NH4)2SO4), can be added at high concentrations to disrupt electrostatic interactions. The ions in these salts can shield the charged amino acid residues, reducing the strength of electrostatic interactions.

Water and heat, mentioned in the question, are not typically used to specifically dampen out electrostatic interactions among amino acid residues. Water is a medium in which proteins naturally exist, and heat can denature proteins by disrupting the weak interactions that stabilize their structure. However, these factors do not specifically target electrostatic interactions.

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Answer:

0.96 J/g °C

General Formulas and Concepts:

Thermochemistry

Specific Heat Formula: q = mcΔT

Explanation:

Step 1: Define

[Given] m = 5.0 g

[Given] ΔT = 22.0 °C - 20.0 °C = 2.0 °C

[Given] q = 9.6 J

[Solve] c

Step 2: Solve for c

Answer: C

Explanation: They both contain membrane-bound organelles such as the nucleus, mitochondria, endoplasmic reticulum, Golgi apparatus, lysosomes, and peroxisomes.

The percent of the mass in a sample of zinc sulfate that comes from zinc would be 40.50%. Option 5.

The percent composition of a component in a substance is the ratio of the mass of the component and the mass of the substance itself.

This can be mathematically expressed as;

Pecent composition= mass of component/mass of substance x 100%

In this case, the substance is zinc sulfate, a compound that contains zinc, sulfur, and oxygen in a ratio 1:1:4.

The molar weight of zinc is 65.38 while the molar mass of the entire zinc sulfate is 161.442.

Thus, the percent of the mass of any sample of zinc sulfate that comes from zinc would be:

Percent zinc = 65.38/161.442 x 100

                    = 40.50%

In other words, the percent of zinc in any zinc sulfate sample would be 40.50%

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If the plate you selected had only been inoculated with 0.1 ml of the dilution, it would have less bacterial colonies and would affect your experiment's results.Inoculation is the process of introducing a sample of bacteria into a culture medium to propagate it.

It entails the injection of bacterial cultures into the experimental environment to monitor their growth and conduct research on them.What happens when a plate is inoculated with a smaller amount of bacteria?A lower bacterial count may occur on the plate if a smaller amount of bacteria is inoculated. When this occurs, the experiment's results may be affected. If the initial number of bacteria is smaller, the number of colonies on the agar plate may be lower. It's critical to get the proper inoculum concentration on the agar plate to obtain accurate results.

How to prevent the plate from being inoculated with a small amount of bacteria?To ensure that the bacteria are properly inoculated, a standard operating procedure must be established. For example, before inoculating the bacteria into the media, the culture should be mixed well and diluted properly. Then, you can obtain a good quantity of bacteria in the inoculum loop, dip it into the broth culture, and streak it onto the agar plate.  As a result, you will be able to obtain an optimal inoculum size. This ensures that the bacteria are evenly distributed over the plate and that they have sufficient room to grow.

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Student 1 is correct. He discovered the mistake as;

If the spot was placed below the solvent level, this would cause the sample to be dissolved into the solvent pool before traveling up the plate.

TLC should be positioned so that just the lowest edge of the plate touches the solvent.

Differential affinities (strength of adhesion) of the analyte components towards the stationary and mobile phases result in differential separation of the components. Affinity, in turn, is determined by two molecular properties: adsorption and solubility.

Adsorption is the property of how well a component of a mixture clings to the stationary phase, whereas solubility is the quality of how well a component of a mixture dissolves in the mobile phase.

The greater the adsorption to the stationary phase, the slower the molecule will move across the column.

The more the molecule adsorbs to the stationary phase, the slower it moves across the column.

The greater the molecule's solubility in the mobile phase, the faster it will travel across the column.

Thus, Student 1 is correct. He detected the error by placing the area below the solvent level, which caused the sample to disintegrate into the solvent pool before migrating up the plate.

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Answer:

An element is a single atom by itself found on the periodic table. A compound  is made when two or more elements chemically bond together. An example of an element is hydrogen. (H) An example of a compound is \(H_{2} O\).

Explanation:

explained above

The location and light reflected on the moon throughout the month changes thanks to the moon's lunar phases :

At the beginning of the lunar cycle, the Moon is located between the Sun and the Earth, and is not visible from Earth because it is not reflecting any sunlight. As the Moon moves away from the Sun, a thin crescent shape becomes visible, and it is illuminated on the right-hand side. This phase is called the Waxing Crescent.

The Moon continues to move away from the Sun, and more than half of its surface becomes illuminated. This is called the Waxing Gibbous phase. After two weeks, the Moon is located on the opposite side of the Earth from the Sun, and its entire visible surface is illuminated. This is called the Full Moon phase.

The Moon continues to move closer to the Sun, and less than half of its surface is illuminated. This is called the Waning Crescent phase.

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The p-value is greater than the significance level of 0.05, we do not reject the null hypothesis and conclude that all proportions are equal.

Firstly, let us conduct a Chi-square test of independence of categorical variables based on the information given above. We have three different cases of hypothesis testing that we have to solve one by one.

Case 1: HD​:pSun ​=rhoMan ​=pTue ​=pWed ​=pThu ​=pFri ​=pSat ​=71​
Ha​ : Not all proportions are equal.

Test Statistic
For this hypothesis, we need to compute the test statistic that is given as:
\($$\chi^2=\sum_{i=1}^{k}\frac{(O_i-E_i)^2}{E_i}$$\)  where k is the number of groups/categories. Since we have 7 days of the week, \(k = 7. $O_i$ and $E_i$\) are the observed and expected frequencies respectively.  
Here, we have equal proportions of 71 for each day of the week.
Therefore, the expected frequencies are also equal to 71.
\($$E_i = 71, i=1,2,3,4,5,6,7.$$\)

We also have to use the given information to compute the observed frequencies,
\($O_i$.$$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126$$\)

Therefore, the test statistic can be computed as \($$\chi^2=\frac{(90-71)^2}{71} + \frac{(99-71)^2}{71} + \frac{(122-71)^2}{71} + \frac{(123-71)^2}{71} + \frac{(130-71)^2}{71} + \frac{(160-71)^2}{71} + \frac{(126-71)^2}{71}$$$$\chi^2=180.14\)

Now we have to find the p-value of this test. Since the number of degrees of freedom is k - 1 = 7 - 1 = 6, the p-value can be found using the chi-square distribution table with 6 degrees of freedom at the 0.05 significance level. The p-value is 0.000014. ConclusionSince the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that not all proportions are equal.

The total number of accidents is \($$90+99+122+123+130+160+126=850$$\)

The percentage of accidents occurring on each day of the week can be found as follows:
\($$Sunday: $$\frac{90}{850}\times 100 = 10.59\%$$Monday: $$\frac{99}{850}\times 100 = 11.65\%$$Tuesday: $$\frac{122}{850}\times 100 = 14.35\%$$Wednesday: $$\frac{123}{850}\times 100 = 14.47\%$$Thursday: $$\frac{130}{850}\times 100 = 15.29\%$$Friday: $$\frac{160}{850}\times 100 = 18.82\%$$Saturday: $$\frac{126}{850}\times 100 = 14.82\%$$\)

From the above percentages, we can see that Friday has the highest percentage of traffic accidents.

Case 2:
HD​: Not all proportions are equal.

Ha​:pSun ​=pMon ​=pTue ​=rhoWed ​=pThu ​=rhoFri ​=rhoSat ​=71

Test Statistic

\($$E_1 = 78.57, E_2 = 86.57, E_3 = 106.86, E_4 = 107.43, E_5 = 113.57, E_6 = 139.43, E_7 = 109.14$$\)

We already know the observed frequencies,
\($$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126.$$\)

The test statistic can be computed as:

\($$\chi^2=\frac{(90-78.57)^2}{78.57} + \frac{(99-86.57)^2}{86.57} + \frac{(122-106.86)^2}{106.86}+ \frac{(123-107.43)^2}{107.43} + \frac{(130-113.57)^2}{113.57} + \frac{(160-139.43)^2}{139.43} + \frac{(126-109.14)^2}{109.14} $$$$ \implies \chi^2=34.98$$\)
The p-value is 0.000001.
Conclusion- Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that not all proportions are equal.

Case 3:

All proportions are equal.
Test Statistic
The expected frequency for each group is
\(E = \frac{850}{7} = 121.43\)
We already know the observed frequencies,
\($$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126.$$\)

The test statistic is,

\($$\chi^2=\frac{(90-121.43)^2}{121.43} + \frac{(99-121.43)^2}{121.43} + \frac{(122-121.43)^2}{121.43} + \frac{(123-121.43)^2}{121.43} + \frac{(130-121.43)^2}{121.43} + \frac{(160-121.43)^2}{121.43} + \frac{(126-121.43)^2}{121.43}} \\\implies \chi^2=9.17$$\)

The p-value is 0.1664.

Since the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis and conclude that all proportions are equal.

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Answer

Explanation:

he most recent Ebola virus outbreak in West Africa, which was unprecedented in the number of cases and fatalities, geographic distribution, and number of nations affected, highlights the need for safe, effective, and readily available antiviral agents for treatment and prevention of acute Ebola virus (EBOV) disease (EVD) or sequelae1. No antiviral therapeutics have yet received regulatory approval or demonstrated clinical efficacy. Here we report the discovery of a novel small molecule GS-5734, a monophosphoramidate prodrug of an adenosine analogue, with antiviral activity against EBOV. GS-5734 exhibits antiviral activity against multiple variants of EBOV and other filoviruses in cell-based assays. The pharmacologically active nucleoside triphosphate (NTP) is efficiently formed in multiple human cell types incubated with GS-5734 in vitro, and the NTP acts as an alternative substrate and RNA-chain terminator in primer-extension assays using a surrogate respiratory syncytial virus RNA polymerase. Intravenous administration of GS-5734 to nonhuman primates resulted in persistent NTP levels in peripheral blood mononuclear cells (half-life, 14 h) and distribution to sanctuary sites for viral replication including testes, eyes, and brain. In a rhesus monkey model of EVD, once-daily intravenous administration of 10 mg kg−1 GS-5734 for 12 days resulted in profound suppression of EBOV replication and protected 100% of EBOV-infected animals against lethal disease, ameliorating clinical disease signs and pathophysiological markers, even when treatments were initiated three days after virus exposure when systemic viral RNA was detected in two out of six treated animals. These results show the first substantive post-exposure protection by a small-molecule antiviral compound against EBOV in nonhuman primates. The broad-spectrum antiviral activity of GS-5734 in vitro against other pathogenic RNA viruses, including filoviruses, arenaviruses, and coronaviruses, suggests the potential for wider medical use. GS-5734 is amenable to large-scale manufacturing, and clinical studies investigating the drug safety and pharmacokinetics are ongoing.

Main

The 2013–2016 outbreak of EVD in West Africa was the largest and most complex EBOV outbreak in the recorded history of the disease, with >28,000 EVD cases and >11,000 reported deaths1. Medical infrastructures in Guinea, Sierra Leone, and Liberia were seriously impacted by a loss of >500 healthcare workers1. Additionally, EVD-related sequelae (joint and muscle pain, as well as neurological, ophthalmic, and other symptoms) together with viral persistence and recrudescence in individuals who survived the acute disease have been documented2,3,4,5.

EBOV is a single-stranded negative-sense non-segmented RNA virus from the Filoviridae family. In addition to EBOV, other related viruses, namely Marburg, Sudan, and Bundibugyo viruses, have caused outbreaks with high fatality rates6. Although the efficacy of various experimental small molecules and biologics have been assessed in EVD animal models and in multiple clinical trials during the West African outbreak7,8,9,10,11,12,13,14,15,16,17,18, there are no therapeutics for which clinical efficacy and safety have been established for treatment of acute EVD or its sequelae. The availability of broadly effective antiviral(s) with a favourable benefit/risk profile would address a serious unmet medical need for the treatment of EBOV infection.

A 1′-cyano-substituted adenine C-nucleoside ribose analogue (Nuc) exhibits antiviral activity against a number of RNA viruses19. The mechanism of action of Nuc requires intracellular anabolism to the active triphosphate metabolite (NTP), which is expected to interfere with the activity of viral RNA-dependent RNA-polymerases (RdRp). Structurally, the 1′-cyano group provides potency and selectivity towards viral RNA polymerases, but because of slow first phosphorylation kinetics, modification of parent nucleosides with monophosphate promoieties has the potential to greatly enhance intracellular NTP concentrations20. GS-5734, the single Sp isomer of the 2-ethylbutyl L-alaninate phosphoramidate prodrug (Supplementary Information), effectively bypasses the rate-limiting first phosphorylation step of the Nuc (Fig. 1a). In human monocyte-derived macrophages, incubation with GS-5734 caused rapid loading of cells with high levels of NTP that persist with a half-life (t1/2) of 24 h following removal of GS-5734 (Extended Data Fig. 1a), resulting in up to 30-fold higher levels compared to incubation with Nuc (Fig. 1b). In cell-based assays, GS-5734 is active against a broad range of filoviruses including Marburg virus and several variants of EBOV (Fig. 1c). GS-5734 inhibits EBOV replication in multiple relevant human cell types including primary macrophages and human endothelial cells with half-maxi

Approximately 78% of the atmosphere is composed of nitrogen gas.

The envelope of gases surrounding the Earth or another celestial body is referred to as the atmosphere. It is held in place by the planet's gravity. The atmosphere protects life on Earth by absorbing ultraviolet solar radiation, warming the surface through heat retention (greenhouse effect), and reducing temperature extremes between day and night (the diurnal temperature variation).

The atmosphere consists of various gases, including nitrogen, oxygen, and carbon dioxide. The percentage of each gas present in the atmosphere is referred to as its composition. The atmosphere of the Earth is made up of 78.08% nitrogen, 20.95% oxygen, 0.93% argon, and trace amounts of other gases such as carbon dioxide, neon, helium, and methane.

The following are the various layers of Earth's atmosphere:

Exosphere

Thermosphere

Mesosphere

Stratosphere

Troposphere

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Because neurons are full of proteins that tend to be negatively charged, they tend to attract positively charged ions.

The interior of neurons is filled with a variety of proteins that have a negative charge. This negatively charged environment tends to attract positively charged ions, such as sodium (Na+) and potassium (K+) ions. The movement of these ions across the cell membrane and into the interior of the neuron is an important aspect of neuronal signaling and communication.

In neurons, the movement of positively charged ions is regulated by ion channels and pumps, which control the flow of ions in and out of the cell. This regulation is critical for maintaining the resting potential of the neuron and for transmitting signals along the length of the neuron. For example, when an electrical signal reaches the end of a neuron, the ion channels there open, allowing positively charged ions to flow into the cell. This creates a change in the electrical potential of the neuron that triggers the release of neurotransmitters, which can then activate other neurons or target cells.

In summary, the negatively charged environment inside neurons attracts positively charged ions, which play a crucial role in neuronal signaling and communication.

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Answer:

a

Explanation:

I'll explain along side with potassium (k) and Calcium (Ca). the unit for scandium is SC

Sc = 21

Ca = 20

K =19

k = 1s²,2s²,2p⁶,3s²,3p⁶,3d⁰,4s¹

Ca = 1s²,2s²,2p⁶,3s²,3p⁶,3d⁰,4s²

Sc = 1s²,2s²,2p⁶,3s²,3p⁶,3d¹,4s²

if you notice above Sc has 3d¹ while Ca has 3d⁰. This is because Sc is a transition metal

Transition metals are metals whose d-orbital aré partially filled with electron

The molarity of the chloride ion is 2.0 * 10⁻⁵ M.

The molarity of the chloride ion is calculated using the formula below:

The percentage concentration of chloride ion = 727/kg * 10⁻⁶ Kg * 100/1

The percentage concentration of chloride ion = 0.0727%

Molar mass of Chloride ion = 35.5 g/mol

Molarity = (0.0727 * 1.035)/(35.5 * 100)

Molarity of chloride ion = 2.0 * 10⁻⁵ M

In conclusion, the molarity of the chloride ion is obtained from the density and percent concentration of chloride ions in seawater.

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Answer:

1250000

Explanation:

Specific heat is mass times specific heat capacity times change in temperature to get chance in temperature minus the old temperature from the new temperature then complete equation hence 25g×1000j×(60-10)

25g×1000j×50

=1250000g/j/degree Celsius

Answer:

The law of conservation of matter can be stated differently in terms of what happens during a chemical reaction as follows:

Explanation:

During a chemical reaction, matter is neither created nor destroyed, but is instead transformed from one form to another. In other words, the total mass of the reactants (the substances that undergo the reaction) is equal to the total mass of the products (the substances that are formed as a result of the reaction). This means that the number and type of atoms in the reactants must be the same as the number and type of atoms in the products, although their arrangement and bonding may be different. Therefore, the mass of the substances before the reaction must be equal to the mass of the substances after the reaction, since matter cannot be created or destroyed.

Total internal reflection is the entire reflection of a light beam back into a medium, such as glass or water, from the surfaces around it.

The correct option is (a) θ1 must be greater than θc and (b) n2 must be less than n1

If a light beam is incident at an angle greater than simply the critical angle of the media, it will always totally reflect back into the optically denser medium from which it originated. Internal introspection that is comprehensive is what this is.

The following are the two requirements for the complete internal reflection to occur:

The angle of incidence in a denser medium when the angle of refraction within a rarer medium approaches 90° is known as the critical angle.

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I understand that the question you are looking for is "What conditions are necessary for total internal reflection to occur? choose all that apply.

(a) θ1 must be greater than θc

(b) n2 must be less than n1

(c) n2 must be greater than n1

(d) θ1 must be less than θc"

Answer:

B

Explanation:

Graphs are tools that help in performing calculations as they reduce the source of errors which is done in long mathematical calculations.

Errors in chemical analysis result when there is a difference between observed value and the true value.If the magnitude of errors is large , it results in decrease in accuracy, reproducibility, and precision.

There are three types of errors:1) random error 2) systematic error 3) human error.The cause of random errors are difficult to quantify while the human errors can be minimized by taking a range of readings to reduce the error.

Errors while measuring boiling point may be human errors while noting down the boiling temperature or instrumental or systematic error if there is a fault in the thermometer.

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To ensure accurate measurement of potentials in electrochemical cells, it was necessary to change solutions even if they used the same reagents.

Failing to do so could result in decreasing accuracy of the potentials. The accuracy of potential measurements in electrochemical cells relies on the establishment of a well-defined reference electrode potential. When two different solutions with the same reagents are used in consecutive measurements without changing the solutions, the composition of the electrolyte might alter due to various factors such as ion migration, solution contamination, or side reactions.

These changes can lead to a deviation from the desired reference potential and result in less accurate measurements. By changing solutions between cells, any variations in the electrolyte composition are minimized, ensuring that the potentials measured are more reliable and accurate.

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Answer:

it's a segment

Explanation:

it has multiple end points

Answer:

Yes

Explanation:

It is important to learn about in science

Answer : Yes

Explanation:

One reason for the importance of kinetics is that it provides evidence for the mechanisms of chemical processes. Besides being of intrinsic scientific interest, knowledge of reaction mechanisms is of practical use in deciding what is the most effective way of causing a reaction to occur.

Answer:

To get the answer, you want to look at the balanced equation and note that 2 moles of aluminum (Al) can produce 1 mole of aluminum oxide (Al2O3). That's the critical relationship that exists.

Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl

To calculate the concentration of NaCl in the resulting solution in mg/L NaCl, we can use the formula; Concentration (mg/L) = (Mass of solute ÷ Volume of solution in L) × 1000 g / 1 mg NaCl is present in the stock solution of 25 mL. So, the mass of NaCl in the solution would be;0.0842 g ÷ 25 mL = 0.00337 g/mL. Now, in the resulting solution, a 1.00 mL aliquot of the stock solution is transferred to a 50.00 mL volumetric flask and diluted to volume. Therefore, the volume of the resulting solution is 50.00 mL. We will substitute these values in the formula, Concentration (mg/L) = (0.00337 g/mL ÷ 50 mL) × 1000 g / 1 mg concentration (mg/L) = 67.4 mg/L. Therefore, the concentration of NaCl in the resulting solution in mg/L NaCl is 67.4 mg/L.7.  Concentration = 67.4 mg/LTolerance = 4.28 mg/LTotal concentration = 67.4 + 4.28 mg/L = 71.68 mg/LWe round off this value to one decimal place; Total concentration = 71.7 mg/LTherefore, the concentration of NaCl in the resulting solution using propagation of error through the calculation is 67.4+0.4 mg/L.8. Two possible sources of random (indeterminate) error in the experiment are; Errors in temperature measurement. Errors in measurement of water volume. Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl.

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Answer:

Explanation:

The early atmosphere

Scientists believe that the Earth was formed about 4.5 billion years ago. Its early atmosphere was probably formed from the gases given out by volcanoes. It is believed that there was intense volcanic activity for the first billion years of the Earth's existence.

The early atmosphere was probably mostly carbon dioxide, with little or no oxygen. There were smaller proportions of water vapour, ammonia and methane. As the Earth cooled down, most of the water vapour condensed and formed the oceans.

It is thought that the atmospheres of Mars and Venus today, which contain mostly carbon dioxide, are similar to the early atmosphere of the Earth.

Scientists can’t be sure about the early atmosphere and can only draw evidence from other sources. For example, volcanoes release high quantities of carbon dioxide. Iron-based compounds are present in very old rocks that could only have formed if there was little or no oxygen at the time.

Changes in the atmosphere

So how did the proportion of carbon dioxide in the atmosphere go down, and the proportion of oxygen go up?

The proportion of oxygen went up because of photosynthesis by plants.

The proportion of carbon dioxide went down because:

it was locked up in sedimentary rocks (such as limestone) and in fossil fuels

it was absorbed by plants for photosynthesis

it dissolved in the oceans

The burning of fossil fuels is adding carbon dioxide to the atmosphere faster than it can be removed. This means that the level of carbon dioxide in the atmosphere is increasing.

Answer:

Neutral subatomic particle that is a constituent of every atomic nucleus except ordinary hydrogen.

Explanation: