2H₂ + O2 → 2H₂O
If you have 3.31 moles of hydrogen how many grams of oxygen are needed?

Answer:

The molarity of the solution is 1.339 M and the molality is 2.48 m.

Explanation:

To calculate the molarity of the solution, you first need to determine the number of moles of sugar present in the solution. To do this, you can use the formula:

moles = grams / molecular weight

For sugar (sucrose), the molecular weight is 342.3 g/mol. So, the number of moles of sugar present in the solution is:

moles = 46 g / 342.3 g/mol = 0.1339 moles

To calculate the molarity of the solution, you then need to divide the number of moles of solute (sugar) by the volume of the solvent (water) in liters. Since there are 100 ml of water in the solution, the volume of the solvent in liters is 0.100 L. So, the molarity of the solution is:

molarity = moles / liters = 0.1339 moles / 0.100 L = 1.339 M

To calculate the molality of the solution, you need to determine the number of moles of sugar present in the solution and the mass of the solvent (water) in kilograms. Since there are 46 grams of sugar in the solution, the mass of the solvent in kilograms is 100 g - 46 g = 54 g. The mass of the solvent in kilograms is therefore 0.054 kg. The molality of the solution is then calculated using the formula:

molality = moles / kilograms = 0.1339 moles / 0.054 kg = 2.48 m

Hydrogen bromide

Basicity (pKb) ~23

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The volume of copper (II) sulfate in a container is equivalent to 0.650L.

Volume is the three-dimensional measure of space that comprises a length, a width and a height. It is measured in units of cubic centimeters (cm³) in metric, cubic inches or cubic feet in English measurement.

According to this question, a container seems to have about 650 mL of copper(II) sulfate solution in volume. The volume can be converted to litres as follows:

1 millilitre = 0.001 litre

650 millilitres = 0.650 litres

Therefore, 0.650L is the volume of copper II sulfate in litres.

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The structure for 2-methyl-3-propalhex-2-yne can be shown in the image attached.

We know that a compound is composed of atoms that can be found in the compound. For the organic compound, we can see that we can be able to obtain the structure of the compound from the structure.

The compound as we can see is  2-methyl-3-propalhex-2-yne. The structure of the compound must be able to include a double bond as we can clearly see from the image that is attached to this answer.

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The main advantage of the Kelvin scale is that all the temperatures on this scale are positive

The Kelvin scale is a temperature scale that has no negative values.

kelvin scale is a scale of temperature. It was Lord William Thomson Kelvin who discovered Kelvin's scale of temperature.

The melting point of ice on the Kelvin scale is 273 K.

The boiling point of water on the Kelvin scale is 373 K.

The SI unit for measuring temperature is Kelvin which is denoted by the symbol ‘K’.

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Molecular formula: \(C8H24O8\) if a 0.5438g sample of liquid contains C, H, and O burned in pure O2 and obtained 1.039g Co2 and 0.6369g H2O. The molar mass of the compound is 376g/mol

To find the empirical formula, we need to determine the ratios of the different elements in the compound. We can do this by finding the number of moles of each element in the sample, using the masses of the products (carbon dioxide and water) and their molar masses.

First, let's find the number of moles of carbon dioxide and water produced:

Number of moles of carbon dioxide= mass of carbon dioxide / molar mass of carbon dioxide

= 1.039 g / 44.01 g/mol

= 0.0236 mol

Number of moles of \(H2O\)= mass of \(H2O\)/ molar mass of \(H2O\)

= 0.6369 g / 18.015 g/mol

= 0.0354 mol

Next, we can use the stoichiometry of the combustion reaction to determine the number of moles of C and H in the sample. The balanced equation for the combustion of a generic hydrocarbon is:

C_aH_bO_c + (a+b/4)O2 → aCO2 + b/2 H2O

Comparing this to the given information, we see that for every 1 mole of \(CO2\)produced, there must be 1 mole of C in the original sample, and for every 1/2 mole of \(H2O\)produced, there must be 1/2 mole of H in the original sample. Therefore:

Number of moles of C = number of moles of carbon dioxide = 0.0236 mol

Number of moles of H = 2 x number of moles of  water= 0.0708 mol

Finally, we can use the number of moles of each element to calculate their ratios in the compound:

C: 0.0236 mol / (0.0236 mol + 0.0708 mol) = 0.25

H: 0.0708 mol / (0.0236 mol + 0.0708 mol) = 0.75

The empirical formula is therefore (assuming the empirical formula has a whole-number ratio of atoms).

To find the molecular formula, we need to determine the molecular mass of the empirical formula and divide the given molar mass of the compound by it:

Molecular mass = (12.011 + 3 x 1.008 + 15.999) g/mol = 47.049 g/mol

Molecular formula = empirical formula x n, where n is an integer

n = molar mass of compound / molecular mass of an empirical formula

= 376 g/mol / 47.049 g/mol

= 7.99

Since n is close to 8, we can round up to get the molecular formula: \(C8H24O8\).

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Answer:

565.4 K

Explanation:

Remember PV = nRT

630 * 50 =  25/28 * 62.4 T

T = 565.4 kelvin

The molarity of the diluted solution is 0.33 M

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 0. 5 M

Volume of stock solution (V₁) = 100 mL

Volume of diluted solution (V₂) = 100 + 50 = 150 mL

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

0.5 × 100 = M₂ × 150

50 = M₂ × 150

Divide both side by 150

M₂ = 50 / 150

Therefore, the molarity of the diluted solution is 0.33 M

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If a gas is at a pressure of 180 Kpa and a temperature of 300 K, the temperature will be 460 K at that pressure.

Gay-Law Lussac's states that, for a given volume, the relationship between pressure and temperature is one of direct proportionality. The initial pressure and temperature of a gas are P1 and T1, respectively. The final temperature and pressure of a gas are P2 and T2, respectively.

According to this, this is. P1 being equal to 180 kpa T2 =? and T1 = 300 K, 276 Kpa, and? replace with the aforementioned formula.

As a result, the offered questions are at a temperature of 460K.

If a gas is at a pressure of 180 Kpa and a temperature of 300 K, the temperature will be 460 K at that pressure.  

The complete question is,

At a temperature of 300 K, a gas has a 180 kPa pressure. How hot will the gas be before it reaches 276 kPa in pressure? Nothing alters the volume.

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The effect of botulinum toxin cannot be determined using a yeast.

Botulinum toxin is produced by the bacterium called Clostridium botulinum. It is a neurotoxin that affects the nervous system. It is done by inhibiting a chemical called as acetylcholine.

Acetylcholine is a neurotransmitter that carries the message from the brain. It relays the message thus resulting in movement. If it is inhibited, the person would be paralyzed.

Yeast is a single-celled organism. It does not have well-developed systems as in multicellular organisms. In addition to that, the test organisms used to measure the effect of toxicity should be similar genetically to humans which would result in higher efficiency.

Thus, yeast cannot be used to determine the effect of botulinum toxin

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Answer:

Given:

Volume of wet hydrogen gas (Vwet) = 4.00 L

Pressure (P) = 737 torr

Temperature (Twet) = 23.0°C = 296 K

To find:

Volume of dry hydrogen gas (Vdry) at STP (Tdry = 273 K and Pdry = 760 torr)

Solution:

Convert temperature to Kelvin: T = Twet = 23.0°C + 273 = 296 K

Use the combined gas law: P1V1/T1 = P2V2/T2

Plug in values: P1 = P, V1 = Vwet, T1 = T, P2 = Pdry = 760 torr, T2 = Tdry = 273 K

Solve for V2 (Vdry): Vdry = Vwet x Pdry x T / P x Tdry

Substitute values: Vdry = 4.00 L x 760 torr x 296 K / 737 torr x 273 K

Simplify: Vdry ≈ 3.48 L

Therefore, the volume of dry hydrogen gas at STP is approximately 3.48 L.

So the answer key is correct, but their method may have been incorrect.

Answer:

The empirical formula of dimethylmercury is C2H6Hg

Explanation:

Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.

At pOH value  of 6.0 the pH value of the following solution is 8.0 and the concentration of [\(H^{+}\) ] ion is \(10^{-8}\)

In this question we will apply the formula

      pH +pOH = 14 . . . . . . . . . . . . .(1)

where pH = concentration of [\(H^{+}\) ] ion

pOH = concentration of [\(OH^{-}\) ] ion

As per the question

pOH =6.0

Putting the value of pOH in equation (1) we get the value of pH

              pH + 6.0 =14

             pH = 14 -6.0

             pH  = 8.0

 The value of pH if the pOH value is 6.0 is 8.0

To find the concentration of \(H^{+}\) ion we will use the following formula

This is calculated by the formula

                                 [\(H^{+}\)} = \(10^{-pH}\)

where we will write the values of pH

Hence the concentration of [\(H^{+}\)} ion is \(10^{-8}\)

Therefore at  pOH of 6.0 the pH value of the following solution is 8.0 and the concentration of [\(H^{+}\) ] ion is \(10^{-8}\)

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The complete question is -

What is the pH value and concentration of [\(H^{+}\) ] ion of the following if the pOH value of the solution is 6.0 ?

Answer:

b. 0.407

Explanation:

Acetic acid / Acetate ion is a buffer (Mixture of a weak acid, acetic acid, with its conjugate base, acetate ion) with pKa = -log Ka = 4.74.

The simplest way to determine the pH of a buffer is using Henderson-Hasselbalch formula:

pH = pKa + log [Conjugate base] / [Weak acid]

For acetic buffer with pH = 4.35:

4.35 = 4.74 + log [A⁻] / [HA]

-0.39 = log [A⁻] / [HA]

0.407 = [A⁻] / [HA]

Thus, right option is:

Answer:

Two of them are solids, one is liquid. Two of them are edible, one is not. One is a mixture, and two are not.

Explanation:

Answer:

chlorine

Explanation:

Answer:

≅ 0.240 grams (3 sig. figs.)

Explanation:

Rxn:   CaCO₃  +          2HCl              => CaCl₂ + H₂O + CO₂

Given:    ?g          45.67ml(0.105M)

                            = 0.04567L x 0.105 mole/L

                            =  0.0048 mole HCl

Rxn ratio for CaCO₃ to HCl is 1:2

∴ moles CaCO₃ consumed = 1/2 of moles HCl used

=> 1/2(0.0048)mole CaCO₃ used = 0.0024 mole CaCO₃

mass CaCO₃ = 0.0024 mole CaCO₃  x  100.09 grams CaCO₃/mole CaCO₃

= 0.23998 grams CaCO₃ (calculator answer)

≅ 0.240 grams (3 sig. figs.)

Answer:

To calculate the heat released in this reaction, we need to use the formula:

Q = mcΔT

where Q is the heat released, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.

Assuming the density of the solution is 1 g/mL, the mass of the solution is 100 g (50 mL HCl + 50 mL NaOH). The specific heat capacity of the solution can be assumed to be the same as that of water, which is 4.18 J/g°C.

The change in temperature can be calculated as the final temperature minus the initial temperature:

ΔT = 30°C - 22°C = 8°C

Therefore, we have:

Q = (100 g) * (4.18 J/g°C) * (8°C) = 3344 J

The heat released in the reaction is 3344 J.

The value of ΔH for the reaction can be calculated using the formula:

ΔH = -Q/n

where Q is the heat released, and n is the number of moles of limiting reactant used in the reaction. In this case, the limiting reactant is NaOH, and we can calculate the number of moles of NaOH from its concentration and volume:

n(NaOH) = (0.1 L) * (1 mol/L) = 0.1 mol

Therefore, we have:

ΔH = -(3344 J) / (0.1 mol) = -33,440 J/mol

The value of ΔH for the reaction is -33,440 J/mol, which is negative because the reaction is exothermic (heat is released).

One example of decay series is the sequence of decays that starts with Thorium-232 which ends up as Lead-208.

What does it mean?

We mean that Thorium-232 decays into Radium-228 plus and an alpha particle. Then, Radium subsequently decays into Actinium-228 via beta decay, and so on until Lead-208 is the final end product.

Answer: Radium-228

Answer:

0.521 moles still present in the container.

Explanation:

It is possible to answer this question by using the general gas law, that is:

PV = nRT

Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)

Replacing with values of the initial conditions of the container, its volume is:

V = nRT / P

V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm

V = 12.9L

When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and the volume of the container still constant. Again, using general gas law:

PV / RT = n

0.998atm*12.9L / 0.082atmL/molK*301.25K = n

0.521 moles = n

Ideal gas law is the hypothetical equation in which the pressure, volume, and temperature of the gas are directly related. It can be denoted as:

PV = nRT

The number of moles still present in the container is 0.521.

The ideal gas law is:

PV = nRT

where,

P = Pressure

V = Volume

R = Gas constant

T =Temperature

n = moles

Given:

Moles in container = 2.00

Temperture = 294.85 K

Pressure = 3.75 atm

Volume =?

Substituting the values:

V = nRT/P

\(\text V&=\dfrac{2\times0.082 \times294}{3.75}\)

Volume = 12.9 L

Now, the condition when changed, such that temperature is 301.25 K, pressure is 0.998 atm, and Volume is 12.9 L, then moles will be equal to:

\(\begin{aligned}\dfrac{\text{PV}}{\text{RT}}&=\text n\\\\\dfrac{0.998 \times 12.9} {0.082 \times 301.25}&=\text n\end\)

n = 0.521 moles

Therefore, 0.521 moles is still present in the container.

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Answer:

\(2,092,000J\)

Explanation:

Hello.

In this case, we need to perform a unit conversion problem between cal and joule, taking into account that 1 cal equals 4.184 joules of energy, therefore, we write the following proportional factor:

\(500,000cal*\frac{4.184J}{1cal}\)

Whereas the 1 cal is written under the line in order to simplify the initial can and the 4.184 J on the line as joules are the required units. Thus, we obtain:

\(2,092,000J\)

As the energy required to burn off such amount of energy given of potato chips.

Best regards.

The enthalpy change of the reaction C(diamond) → C(graphite) is -2.9 kJ.

The given information is ΔHrxn for the reaction C(diamond) → C(graphite) can be calculated with the given equations:Equations:  C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ  2 CO2(g) → 2 CO(g) + O2(g) ΔH = 566.0 kJ  C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ  2 CO(g) → C(graphite) + CO2(g) ΔH = −172.5 kJThe required reaction can be obtained by adding the equations (1) and (4), as follows:C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)Addition of the two equations (1) and (4) results in a reaction whose products are C(graphite) and CO2.

To get the final equation that involves only the required reactants and products, the equation (2) should be added, which consumes CO2 and produces O2, as shown below:C(diamond) + O2(g) → CO2(g)  ΔH = −395.4 kJ [eq. (1)]  2 CO(g) → C(graphite) + CO2(g)  ΔH = −172.5 kJ [eq. (4)]  2 CO2(g) → 2 CO(g) + O2(g)  ΔH = 566.0 kJ [eq. (2)]  C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)  ΔHrxn=ΣΔHf(products)−ΣΔHf(reactants)  ΔHrxn=[(3 mol CO2)(-393.5 kJ/mol) + (1 mol C(graphite))(0 kJ/mol)] − [(1 mol C(diamond))(0 kJ/mol) + (1 mol O2)(0 kJ/mol) + (2 mol CO(g))(−172.5 kJ/mol)] − [(2 mol CO2)(566.0 kJ/mol)]  ΔHrxn=−2.9 kJ.

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By analyzing the potential energy diagrams and calculating ΔH using bond energies, we can determine whether a reaction is exothermic (ΔH < 0) or endothermic (ΔH > 0).

To create potential energy diagrams for chemical reactions and determine the value of ΔH (the change in enthalpy) for each reaction, we need to understand the basic concepts and steps involved.

Potential Energy Diagram: A potential energy diagram is a graphical representation of the energy changes that occur during a chemical reaction. The vertical axis represents the potential energy, while the horizontal axis represents the progress of the reaction.

Reactants and Products: Identify the reactants and products involved in each reaction. Assign them appropriate labels on the potential energy diagram.

Activation Energy: Determine the activation energy (Ea) for each reaction. It represents the energy barrier that must be overcome for the reaction to occur. On the diagram, the reactants' energy level is typically higher than the products' energy level, with the activation energy peak in between.

Transition State: Locate the highest point on the potential energy diagram, which represents the transition state or activated complex. This point indicates the highest energy level during the reaction.

ΔH Determination: ΔH represents the difference in enthalpy between the reactants and products. It can be determined by examining the vertical distance between the reactants' energy level and the products' energy level on the potential energy diagram.

ΔH Calculation: ΔH can be calculated using the formula ΔH = Σ (bond energies of reactants) - Σ (bond energies of products). The bond energies are the energy required to break a particular bond or released when a bond is formed.

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Answer:

This might be a scam because it is a common tactic used by scammers to evoke sympathy from their targets. By appealing to the target's emotions, they may be more likely to overlook any warning signs and take action without thinking. Additionally, the urgency of the request (i.e., the tickets could go to waste) may pressure the target to act quickly without considering the potential risks. Scammers may use similar tactics to obtain personal information or money from their targets, so it's always important to be cautious when receiving unsolicited requests.

Explanation:

Answer:

Iron is being oxidised.

Explanation:

To determine the right answer to the question, it is important that we write the half reaction.

The equation for the half reaction is given below:

Fe —> Fe^2+ + 2e^-

Next, we shall determine the change in the oxidation number of iron, Fe. This is illustrated below:

Fe = 0

Fe = +2

From the above, we can see that the oxidation number of iron, Fe changes from 0 to +2 i.e the oxidation number of iron, Fe increased.

This indicates that iron, Fe is being oxidised.

Answer:

B. iron is being oxidized

Explanation:

I got it right

Answer:

protons and electrons

Explanation:

because two atoms of the same elements have different(mass)isotope so the answer is B

21.1 grams of calcium chloride will be produced when 27.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid.

The reactant that is consumed first in a chemical reaction is the limiting reagent or limiting reactant because it stops any more reactions from taking place. The limiting reagent controls how much product is produced during the reaction.

CaCO3(s) + 2HCl(aq) ==> CaCl2(aq) + H2O(l) + CO2(g)  

\(mols CaCO3 =\) \(27 *\frac{1 mol}{100g}\) = 0.27 mol

\(mols HCl = 15 gx \frac{1 mol}{36.5 g}\) = 0.38 mol

Limiting reactant = HCl ( it takes 2x the mols of HCl compared to mols CaCO3 as per balanced equation)

mass of CaCl2 = \(0.411 mols HCl x\frac{ 1 mol CaCl2}{ 2 mols HCl } x 111.0 g/mol\)

mass of CaCl2 = 21.1 g

Hence, 21.1 g of calcium chloride will be produced when 27.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid.

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The mass of Aluminum Sulfate is 5.373 grams if you have \(2.837*10^{26\) atoms of Sulfur .

The molecular formula of Aluminum Sulfate is \(Al_2(SO_4)_3.\) In one molecule of aluminum sulfate, there are 3 sulfur atoms. To calculate the mass of aluminum sulfate, follow the steps below:

Step 1: Calculate the molar mass of aluminum sulfate using the periodic table.Al = 27.0 g/molS = 32.1 g/molO = 16.0 g/mol

(2 × Al) + (3 × S) + (12 × O) = molar mass of \(Al_2(SO_4)_3.\) = 342.2 g/mol

Step 2: Find the number of moles of sulfur in the given number of atoms of sulfur.2\(2.837*10^{26\) atoms of sulfur × 1 mol S/\(6.022 * 10^{23\)atoms S = 0.0470 mol S

Step 3: Use the molar ratio of sulfur to aluminum sulfate to calculate the number of moles of aluminum sulfate.1 mol \(Al_2(SO_4)_3.\) / 3 mol S = 0.333 mol\(Al_2(SO_4)_3.\) per mol S0.0470 mol S × 0.333 mol \(Al_2(SO_4)_3.\)/mol S = 0.0157 mol \(Al_2(SO_4)_3.\)

Step 4: Calculate the mass of aluminum sulfate.0.0157 mol \(Al_2(SO_4)_3.\) × 342.2 g/mol\(Al_2(SO_4)_3.\)= 5.373 g\(Al_2(SO_4)_3.\)

Therefore, the mass of Aluminum Sulfate is 5.373 grams if you have \(2.837*10^{26\) atoms of Sulfur.

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