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Drupady [299]
2 years ago
10

Oxygen can be produced in the laboratory by the reaction

Chemistry
1 answer:
Marina CMI [18]2 years ago
3 0

Answer:

5.2g potassium chlorate is needed to do so

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2H2S + 3O2 → 2SO2 + 2H2O<br><br><br>1. For every 10.0 g of H2S how many moles of SO2 are formed?
Veseljchak [2.6K]

Answer:10

Explanation:

7 0
2 years ago
Calculate the number of moles of an ideal gas if it occupies 1750 dm3 under 125,000 pa at a temperature of 127 c. a. 0.21 moles
labwork [276]

The number of mole will be 65.81 mole.

An ideal gas would be one for which both the overall volume of the molecules and even the forces that exist between them are so negligible as to have no influence on the behavior of something like the gas.

Number of ideal gas can be calculated by using the formula:

PV = nRT

where, p is pressure, n is number of mole, R is gas constant and T is temperature.

Given data:

V= 1750 dm^{3} = 1750 L

P = 125,000 p = 1.2 atm

R = 0.082 L /mole kelvin

T = 273+127 = 400 K

Now, put the value of given data in above equation.

1.23atm x 1750L = n x 0.0820atm x Liter/ mole x kelvin  x 400K

n = 65.81 mole.

Therefore, the number of mole will be 65.81 mole

To know more about mole

brainly.com/question/21050624

#SPJ4

6 0
2 years ago
How many grams are in 1.76 x 10^23 atoms of iodine
Mariana [72]

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

5 0
3 years ago
If two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then
topjm [15]

Answer:

Higher pressure, is the right answer.

Explanation:

The A will have a higher pressure. Since we have given the volume and temperature is same in both containers A and B. Below is the calculation for proof that shows which container has the higher pressure while keeping the volume and temperature the same.  

So, \ V_A = V_B \\\frac{n_A T_A}{P_A} = \frac{n_B T_B}{P_B} \\Here, \ T_A = T_B \\P_A = \frac{n_A}{n_B} \times P_B \\\frac{n_A}{n_B} > 1 \\\frac{P_A}{P_B} > 1 \\P_A > P_B \\

Therefore, the container “A” will have higher pressure.

8 0
2 years ago
The following balanced equation shows the formation of sulfur dioxide.
AlexFokin [52]

actually the answer is C. 15.0 on ed

6 0
3 years ago
Read 2 more answers
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