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3 years ago

The mass of 322 mL of hydrogen gas at 4.0 atm and –73°C is 1.21 grams. What is the density of hydrogen gas at STP?

1 answer:

7 0

To be able to answer this item, we assume that the given hydrogen gas is ideal such that we are able to use the Ideal Gas equation,
PV = nRT
At STP, the values of volume, pressure, and temperature are 22.4 L, 1 atm, and 273.15 K. Solving for n,
n = (1 atm x 22.4 L) / (0.0821 L.atm/mol K x 273.15 K)
n = 0.9988 mols
Each mol of hydrogen gas is 2 g.
m = (0.9988 mols) x (2 g/1 mol)
m = 1.9977 g
Density is the quotient of mass and volume,
density = 1.9977 g/ 22400 mL
density = 8.92 x 10^-5 g/mL

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3 years ago

Chemistry help!!

Scrat [10]

The student originally has 252 grams of water in this experiment.

LAW OF CONSERVATION OF MASS:

The law of conservation of mass explains that matter (mass) can neither be created nor destroyed but can be changed from one form to another.

This means that in a chemical reaction, the sum of the masses of the reactants must equate to the total mass of product(s).

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Since hydrogen and oxygen are the constituent elements of water, the sum of their masses must equate the mass of water.

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2 years ago

You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level

snow_lady [41]

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 23.2 g

volume = final volume of water - initial volume of water

volume = 62 - 55 = 7 mL

We have

$density = \frac{23.2}{7} \\ = 3.314285$

We have the final answer as

Hope this helps you

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3 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

4 0

2 years ago