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25 days ago

Calculate the grams of oxygen in 90.0 g of cl2o

2 answers:

4 0

To find the mass of oxygen in the specified compound, we require the molar mass for both the compound and oxygen. We also establish the relationship between the number of moles of oxygen per mole of the substance. The calculation proceeds as follows:

90.0 g ( 1 mol / 86.91 g ) ( 1mol O / 1 mol Cl2O) ( 16 g / 1 mol ) = 16.57 g O

Anarel [2.6K]25 days ago

4 0

Answer:

The precise answer is 16.576 g of $O_{2}$ .

Explanation:

Initially, we need to compute the molar mass (Mm) of $O_{2}$ and $Cl_{2}O$ .

Mm $O_{2}$ = 32g

Mm $Cl_{2}O$ = 86.9 g

Given that we have 90 g of $Cl_{2}O$ , we can determine the number of moles of $Cl_{2}O$ within that mass.

90 g $Cl_{2}O$ * $\frac{1 mol Cl_{2}O}{86.9 g Cl_{2}O}$ = 1.036 mol $Cl_{2}O$

In the reaction forming $Cl_{2}O$ , 1 mole of $O_{2}$ yields 2 moles of $Cl_{2}O$ :

$O_{2}$ + 2 $Cl_{2}$ -> 2 $Cl_{2}O$

Utilizing the coefficients, we can figure out the moles of $O_{2}$ reacting:

1.036 mol $Cl_{2}O$ * $\frac{1 mol O_{2}}{2 mol Cl_{2}O}$ = 0.518 mol $O_{2}$

Finally, let’s find out how many grams are contained in the reacting moles of $O_{2}$ :

0.518 mol $O_{2}$ * $\frac{32 g O_{2}}{1 mol O_{2}}$ = 16.576 g $O_{2}$

Wishing you a pleasant day!

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A 30.0 mL sample of hydrogen gas (H2) is collected over water at 20.00∘C and has a total pressure of 700.0 torr. The partial pre

Alekssandra [2719]

Answer: The mole fraction of hydrogen gas at 20°C is 0.975

Explanation:

The information provided includes:

Water vapor pressure at 20°C is 17.5 torr

Total pressure at 20°C = 700.0 torr

Hydrogen gas vapor pressure at 20°C = (700.0 - 17.5) torr = 682.5 torr

To find hydrogen gas's mole fraction at 20°C, we utilize Raoult's law, represented by:

$p_{H_2}=p_T\times \chi_{H_2}$

where,

$p_{H_2}$ = pressure of hydrogen gas = 682.5 torr

$p_T$ = total pressure = 700.0 torr

$\chi_{H_2}$ = mole fraction of hydrogen gas =?

Substituting the values into the equation yields:

$682.5torr=700.0torr\times \chi_{H_2}\\\\\chi_{H_2}=\frac{682.5}{700.0}=0.975$

Thus, the mole fraction of hydrogen gas at 20°C equals 0.975

4 0

20 days ago

A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in w

Anarel [2605]

Answer:

The glycerol solution has a molality of 2.960×10^-2 mol/kg.

Explanation:

Calculating the moles of glycerol involves the formula: Moles = Molarity × Volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles.

To find the mass of water, use: Mass = Density × Volume = 0.9982 g/mL × 998.7 mL = 996.90 g, which converts to 0.9969 kg.

The formula for molality is: Molality = Moles of solute/Mass of solvent (in kg) = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg.

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26 days ago

Which change of state is shown in the model?

Alekssandra [2719]

I think the state change illustrated in the diagram is deposition.
Deposition is the transformation of gases into solids without transitioning through a liquid phase. It is the reverse process of sublimation.
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1 month ago

Read 2 more answers

Consider a specific chemical reaction represented by the equation aA + bB → cC + dD. In this equation the letters A, B, C, and D

castortr0y [2743]

The solution to your inquiry is: c = a + b - d. In the chemical equation aA + bB → cC + dD, c may take any value from 1 to however many are needed for the equation to maintain its balance. For instance, if we consider c + d = a + b, we can express c as a + b - d. If we assign values a = 1; b = 3, and d = 2, then c equals 1 + 3 - 2, resulting in c = 2, and so forth.

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8 days ago

At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?

KiRa [2726]

The quantity of fluorine in moles is calculated as 71/19 = 3.74
We also know that at standard temperature and pressure (273 K and 101.3 kPa), one mole of gas occupies 22.4 liters
So, the volume for 3.74 moles at S.T.P is: 3.74 x 22.4
This results in a volume of 83.776 L, which is equivalent to 83,776 mL

Next, applying Boyle's law, which states that for a fixed amount of gas,
PV = constant

We set up the equation P x 6843 = 101.3 x 83776
Solving for P gives us 1,240 kPa

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17 days ago