How many d electrons does I (atomic number
53) possess?
1. 10
2. 5
3. 8
4. 0
5. 20
6. 53
7. 16

Answer:

In order to identify a chemical property,we look for a chemical change.Chemical changes involve chemical reactions and are often observed through a change in colour or odor

Answer:

Corrosion is your answer

Answer:

▷The answer would be

 ▶︎corrosion

Explanation:

~corrosion~ is a natural process that converts a refined metal into a more chemically stable form such as oxide, hydroxide, or sulfide. It is the gradual destruction of materials by chemical and/or electrochemical reaction with their environment

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Isomers are compounds that have the same molecular formula but different arranged differently of the atoms.

For example

So even if they have the same proportion of their atoms, the chemical identity is different

The answer is option B: no because more than one compound can have the same ratio of masses of elements if the atoms are arranged differently.

Answer:

why would you do that...

Answer:

1

Explanation:

if 33 the answer would be 92 but u think then 92 32 1

Expressing the answer in exponential form we get 3⁴.

To simplify the expression (3⁷/(3³), we can apply the properties of exponents. When dividing two exponential expressions with the same base, we subtract the exponents.

In this case, we have 3⁷ divided by 3³, which can be simplified as:

3⁽⁷⁻³⁾

3⁴

Therefore, the simplified expression is 3⁴.

To understand why we subtract the exponents when dividing, we can break down the steps.

The expression 3⁷ represents 3 multiplied by itself seven times:

3 × 3 × 3 × 3 × 3 × 3 × 3.

The expression 3³ represents 3 multiplied by itself three times:

3 × 3 × 3.

When dividing these two expressions, we can cancel out common factors by subtracting the exponents:

(3 × 3 × 3 × 3 × 3 × 3 × 3) / (3 × 3 × 3)

This simplifies to:

3 × 3 × 3 × 3

Which is equivalent to 3⁴.

Thus, the answer in exponential form is 3⁴

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The final volume of the gas is 15.8 L.

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as-

                      PV = nRT

where,

P = Pressure

V = Volume

T = Temperature

n = number of moles

Given,

n = 1

Pressure = 1 atm

W = 1 kJ

Temperature = 77⁰F

γ = 1.4

PV = nRT

The temperature in K is written as -

T = ( 77 - 32 ) 5/9 + 273.15

= 298.15 K

\(w = \frac{nr( T_{1} - T_{2)} }{\pi - 1}\)

T₂ = 250.04 K

The initial volume of the container is -

P₁V₁ = nRT₁

101.32 Pa × V = 1 × 8.314 × 298

V = 0.025 m³

The final volume of the gas is worked out from the equation -

\((\frac{V_{2} }{V_{1} } ) = (\frac{T_{1} }{T_{2} })^{(1.4 - 1)}\)

V = 0.0158 m³ = 15.8 L

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The C-leaving group link and the C-nucleophile bond are broken. Walden inversion, or nucleophile attack from the back, causes configuration inversion.

Chemical processes occur when atoms form or break chemical bonds. Reactants are the substances that begin a chemical reaction, while products are the compounds that are created as a result of the reaction.

The contemporary understanding of chemical processes states that in order for atoms with in reactants to reassemble into products, bonds between their atoms must be broken. Breaking bonds requires energy, and as bonds are broken, energy is evolved.

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The amount of CO2 emitted from the combustion of 1 gallon of gasoline is 8.627 kg.

To calculate the amount of CO2 emitted from the combustion of 1 gallon of gasoline, we need to use the balanced chemical equation for the combustion of octane (C8H18), which is the main component of gasoline:

C8H18 + 12.5O2 → 8CO2 + 9H2O

From this equation, we can see that for every 1 mole of C8H18 that is combusted, 8 moles of CO2 are produced. To find out how many moles of C8H18 are in 1 gallon of gasoline, we need to use the molar mass of C8H18:

Molar mass of C8H18 = (8 x 12.01) + (18 x 1.01) = 114.22 g/mol

Now we can use the weight of 1 gallon of gasoline (2.8 kg or 2800 g) to find out how many moles of C8H18 are in 1 gallon:

Moles of C8H18 = 2800 g / 114.22 g/mol = 24.5 mol

Finally, we can use the balanced chemical equation to find out how many moles of CO2 are produced from the combustion of 24.5 moles of C8H18:

Moles of CO2 = 24.5 mol x 8 = 196 mol

To find out how much CO2 is emitted in grams, we need to use the molar mass of CO2:

Molar mass of CO2 = (12.01) + (2 x 16.00) = 44.01 g/mol

Mass of CO2 = 196 mol x 44.01 g/mol = 8627.96 g or 8.627 kg

Therefore, the amount of CO2 emitted from the combustion of 1 gallon of gasoline is 8.627 kg.

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6.23L volume of h2 gas is produced at stp.

Given that one mole of dihydrogen (H2) weighs 2,0159 g/mol, the experimental density of hydrogen—which is 0,0899 g/L at 0 °C and 0,0837 g/L at 20 °C—can be used to compute the exact molar volume of hydrogen.

This implies that at STP, one mole of hydrogen also takes up 22.4 liters of space. Therefore, at STP, 10moles of hydrogen gas will take up =22.410=224litres of space. In other words, when 460g of salt combines with extra water at STP, 224litres (or 224000mL) of hydrogen gas are released.

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the question you are looking for is

A 5.00 g sample of Al reacts with excess H2SO4. What volume of H2 gas is produced at STP?

2Al(s) + 3H2SO4(aq)→ Al2(SO4)3(aq) + 3H2 (g)

Answer:

0.779

Explanation:

Determine the molecular weight of the hydrocarbon. We know that its vapor density is 29, which means that one mole of the hydrocarbon has a mass of 29 grams. Therefore, the molecular weight of the hydrocarbon is 29 g/mol.

Calculate the number of moles of the hydrocarbon. We can use the formula:

moles = mass / molecular weight

Substituting the values, we get:

moles = 29 g / 29 g/mol = 1 mol

Therefore, we have one mole of the hydrocarbon.

Write the balanced chemical equation for the combustion of the hydrocarbon in oxygen. The general equation is:

hydrocarbon + oxygen → carbon dioxide + water

For one mole of the hydrocarbon, we need one mole of oxygen to completely burn it. The balanced equation is:

CnHm + (n+m/4) O2 → n CO2 + m/2 H2O

Calculate the volume of carbon dioxide produced. We know that 1 mole of any gas at STP occupies 22.4 L. Therefore, one mole of carbon dioxide occupies 22.4 L. The volume of 448 ml of carbon dioxide at STP can be converted to liters:

448 ml = 0.448 L

The number of moles of carbon dioxide produced can be calculated using the ideal gas law:

PV = nRT

where P is the pressure (1 atm), V is the volume (0.448 L), n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature (273 K). Substituting the values, we get:

n = PV/RT = (1 atm x 0.448 L) / (0.0821 L atm/mol K x 273 K) = 0.0177 mol

Therefore, 0.0177 moles of carbon dioxide are produced.

Calculate the mass of carbon dioxide produced. We can use the formula:

mass = moles x molecular weight

The molecular weight of carbon dioxide is 44 g/mol. Substituting the values, we get:

mass = 0.0177 mol x 44 g/mol = 0.779 g

Therefore, the mass of carbon dioxide produced is 0.779 grams.

2. 51 L of a gas at standard temperature and pressure is compressed to 451 mL. The new pressure of the gas is 54 atm.

To find the new pressure of the gas, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Boyle's Law equation: P1V1 = P2V2

Given:

Initial volume (V1) = 51 L

Final volume (V2) = 451 mL = 0.451 L (using the conversion factor: 1 L = 1000 mL)

Initial pressure (P1) = standard pressure = 1 atm

Now we can substitute these values into Boyle's Law equation:

1 atm * 51 L = P2 * 0.451 L

Simplifying the equation:

51 = 0.451 * P2

Dividing both sides of the equation by 0.451:

51 / 0.451 = P2

Calculating:

P2 ≈ 113.3 atm

Therefore, the new pressure of the gas is approximately 113.3 atm.

The new pressure of the gas, when compressed from 51 L to 451 mL, is approximately 113.3 atm.

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There are 3.23 grams of Mg(NO3)2 present in 145 mL of a 0.150 m solution of Mg(NO3)2. To answer this question, we need to use the formula:

moles = concentration x volume

We know the volume of the solution is 145 mL, which is equivalent to 0.145 L. We also know the concentration is 0.150 m. Therefore, we can calculate the number of moles of Mg(NO3)2 in the solution as:

moles = 0.150 mol/L x 0.145 L
moles = 0.02175 mol

Now we need to convert moles to grams. We can do this by using the molar mass of Mg(NO3)2, which is:

Mg(NO3)2 molar mass = (1 x Mg atomic mass) + (2 x N atomic mass) + (6 x O atomic mass)
Mg(NO3)2 molar mass = (1 x 24.31 g/mol) + (2 x 14.01 g/mol) + (6 x 16.00 g/mol)
Mg(NO3)2 molar mass = 148.31 g/mol

Therefore, the mass of Mg(NO3)2 in the solution is:

mass = moles x molar mass
mass = 0.02175 mol x 148.31 g/mol
mass = 3.23 g

So, there are 3.23 grams of Mg(NO3)2 present in 145 mL of a 0.150 m solution of Mg(NO3)2.

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Answer:

See explanation

Explanation:

Number of moles = given mass/molar mass

For Ca(OH)2

n= 31.82g/74g/mol= 0.43 moles

n = mass/ molar mass

Molar mass of K2SO3 = 158g/mol

mass= 3.2 × 158 = 505.6 g

1 formula unit = 6.02 × 10^23

1 mole = 6.02 × 10^23

x moles = 7.25x10^23

x = 7.25x10^23/6.02 × 10^23

x= 1.2 moles

1 mole of Cl2 occupies 22.4 L at STP

x moles of Cl2 occupies 46.6L

x = 46.6/22.4

x= 2.1 moles

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

- log [H⁺] = pH → 6.999

A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

The resulting solution will have a pH of about 4.75 when 25.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH. often known as sodium hydroxide, is a strong base. It's a colorless, odorless substance that's highly hygroscopic.

often known as acetic acid, is an organic acid. It's a weak acid, unlike hydrochloric acid or sulfuric acid. It's a colorless liquid that's highly flammable. It's found in vinegar.What happens when NaOH and CH3COOH are mixed?When NaOH and CH3COOH are combined, they react to create water (H2O), salt, and a weak acid known as CH3COO- (acetic acid ion).This reaction's balanced equation is shown below:CH3COOH + NaOH → CH3COO- Na+ + H2OIn this reaction, the pH of the resulting solution is determined by the concentration of the CH3COOH and CH3COO- ions present. Since CH3COOH is a weak acid, it does not completely dissociate in solution, and some of it remains in its undissociated form, while the rest is dissociated into H+ and CH3COO- ions.The pH of the resulting solution can be calculated using the Henderson-Hasselbalch equation:pH = pKa + log ([A-] / [HA]),wherepKa is the acid dissociation constant for acetic acid, which is 4.76 at 25°C[A-] is the concentration of CH3COO- ions[HA] is the concentration of undissociated CH3COOH ionsWhen 25.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH, the amount of NaOH is not sufficient to completely neutralize all of the CH3COOH in the solution. As a result, there will still be some undissociated CH3COOH in the solution, along with the CH3COO- ions formed as a result of the reaction.The amount of CH3COO- ions generated is the same as the amount of NaOH added, but the amount of undissociated CH3COOH present is determined by the pH of the solution. This leads to a buffer solution being formed, which has a pH near the pKa of acetic acid, which is 4.76.Therefore, when 25.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH, the resulting solution will have a pH of about 4.75.

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Option 1 is correct. There are three types of UV radiation: UVA, UVB, and UVC. UVC has the shortest wavelength, around 100-280 nm, but is mostly absorbed by the Earth's atmosphere and doesn't reach the surface. UVB has a wavelength of 280-320 nm and is responsible for sunburns and skin damage. UVA has the longest wavelength, around 320-400 nm, and can penetrate deeper into the skin, causing aging and wrinkling.

Option 2 is incorrect because UV radiation, in general, can be damaging to the Earth's surface, causing skin cancer, harming plant life, and contributing to climate change.

Option 3 is partially correct because UV radiation with a wavelength of 300 nm falls within the UVC range, but it's important to note that this type of radiation is mostly absorbed by the ozone layer before reaching the Earth's surface.

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The yield of methane from acetic acid is 0.129 g CH4/g CH3COOH.

Methanol is produced using anaerobic digestion of acetic acid. If 0.3 g of CO2 is produced per g acetic acid consumed, the yield of methanol from the substrate (g methanol/g acetic acid) can be calculated as follows:

Solution:

Balanced reaction for anaerobic digestion of acetic acid is:

CH3COOH → CH3COO- + H+ H+ + e- → ½ H2

Hence,CH3COOH → CH3COO- + H+ + ½ H2 For every acetic acid molecule utilized, one methane (CH4) molecule and one carbon dioxide (CO2) molecule are produced.

CH3COOH → CH4 + CO2If 0.3 g of CO2 is produced per g acetic acid consumed, then the yield of methane (CH4) from acetic acid can be calculated as follows:

Yield of methane from acetic acid = (mass of methane produced) / (mass of acetic acid consumed)For each mole of methane produced, one mole of acetic acid is consumed.

So, the molar mass of acetic acid and methane is used to convert between mass and moles.

Yield of methane from acetic acid = (moles of methane produced x molar mass of methane) / (moles of acetic acid consumed x molar mass of acetic acid)

= (0.3 g CO2 / 44.01 g CO2/mol) x (16.04 g CH4/mol) / (1 g CH3COOH / 60.05 g CH3COOH/mol)

= 0.129 g CH4/g CH3COOH

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The total number of valence electrons contained in a sample of 4.78 million kg of sulfur would be 5.3078 x \(10^{32\) valence electrons.

According to Avogadro, 1 mole of any substance contains 6.022 x \(10^{23\) atoms.

4.78 million kg of sulfur = 4,700,800 kg = 4,700,800,000 grams

Mole = mass/molar mass

Mole of 4,700,800,000 grams sample of sulfur = 4,700,800,000/32

= 146900000 moles

146900000 moles = 146900000 x 6.022 x \(10^{23\) atoms

                               = 8.8463 x \(10^{31\) atoms

Now each atom of sulfur contains 6 valence electrons. Thus, total number of valence electrons would be:

6 x 8.8463 x \(10^{31\) = 5.3078 x \(10^{32\) valence electrons

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Answer:

1- Oxygen in water

2- Brass, bronze and sterling silver

3- Air

Answer:

It will be : (in simple terms)

Condensation of hot steam or hot vapor on the cooler skin releases the water's latent heat of vaporization, rapidly raising the temperature of the skin which will result in steam burns.

Hope it helped

All the best!!

Answer:

B holding together molecules in a material

Explanation:

Intermolecular forces are primarily responsible for holding together molecules in a material due to the fact that they are interactive forces between molecules

Answer:

2

Explanation:

Answer:

C

Explanation:

with combustion you must know that your products must be carbon dioxide and water. I think you copied your question incorrectly.

Answer:

If I am not wrong is C x2y5

Explanation:

The concept of valency is used here to determine the formula of the ionic compounds. The number of valence electrons in 'X' is 2 and that of 'Y' is 5. The formula is X₃Y₂. The correct option is D.

The crystalline solids which are formed by neatly packed ions of opposite charges are defined as the ionic compounds. These compounds are formed by the combination of metals with non-metals.

The combining capacity of an element is given by valency. The elements belonging to the same group have the same valency. It is related to how many electrons are in the outer shell.

Here the number of valence electrons in 'X' = 2

The number of valence electrons in 'Y' = 5

The element 'X' need 6 electrons to attain the octet and 'Y' needs 3 electrons to attain the octet.

X - 6

Y - 3

The formula of the ionic compound is X₃Y₂.

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The effect of errors on the calculated concentration of CI is significant.

A 10.00 mL diluted chloride sample was titrated with 0.02749 M AgNO3, and 16.51 mL AgNO, was required to reach the endpoint. The effect of errors on the calculated concentration of CI can be explained as follows:a. The student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M. If the student read the molarity of AgNO, as 0.02479 M instead of 0.02749 M, then the experimentally calculated moles of Ag would be too high. Consequently, the calculated [CI] in the unknown would come out too low. b.

The student was past the endpoint of the titration when the final buret reading was taken. If the student was past the endpoint of the titration when the final buret reading was taken, then the experimentally determined moles of Ag would be too low. This would cause the calculated C1 concentration to come out too high. Consequently, the calculated moles of CI would come out too high. Therefore, the effect of errors on the calculated concentration of CI is significant.

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Answer:

The protons must feel a repulsive force from the other neighboring protons. This is where the strong nuclear force comes in. ... If they can't get that close, the strong force is too weak to make them stick together, and other competing forces (usually the electromagnetic force) can influence the particles to move apart.

Explanation:

(a) At 600 °C, for an Ag-90 at% Sr alloy:

I Draw the phase diagram's constitution point.

(ii) Determine the compositions of the current phases in at%.

(iii) A gradual cooling to 500°C takes place. Will the compositions and ratios of the phases change?

(b) At 600 °C, for an Ag-30 at% Sr alloy:

I Draw the phase diagram's constitution point.

(ii) Recognize the phases and calculate the compositions of each in at%.

(iii) If the temperature is lowered to 500°C, will the proportions change? Why is that so?

(c) The atomic weights of Ag and Sr are 107.9 and 87.6, respectively. Determine the weight percentages of the four intermetallic compounds that make up the Age Sr system.

c) For an Al-4 w t% Copper alloy

I Compute

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Answer:

BURN IT ALIVE MUHAHAHAHAHA

Explanation:

jk

Explanation:

2Al(OH)3 + 3H2SO4 --> Al2 (SO4)3 + 6H2O