If energy absorbed is greater than energy emitted does temperature increase or decrease?

At 273 k and 1.00 x 10^-2 atm, the density of a gas is 1.24 x 10^-5 g/cm3.

bija089 [108]

Root mean square velocity is the square root of the mean of the squares of speeds of different molecules. From kinetic theory of gas, the formula of root mean square velocity=C $_{rms}$ = √ $\frac{3RT}{M}$ =√ $\frac{3PV}{M}$ =√ $\frac{3P}{d}$ , where, R= Universal gas constant, T= Absolute temperature, P= Pressure, V= Volume of gas, d= Density of gas.

Given, T=273 K, P=1.00 x 10⁻² atm, d=1.24 x 10⁻⁵ g/cm³.

(a) Using the formula $C_{rms}$ =√ $\frac{3P}{d}$ =√(3X1.00X10⁻²)/(1.24X10⁻⁵)=49.18

(b) Molar mass can be determined by using the formula $C_{rms}$ =√{3RT}{M}

49.18=√ $\frac{3X8.314X273}{M}$

49.18²=√(3X8.314X273)/M

M= $\frac{3X8.314X273}{49.18^{2} }$

M=1.67 ≅ 2

Molecular mass is 2.

(c) The gas is Helium (He) whose molecular mass is 2.

8 0

3 years ago

Read 2 more answers

Ppppoooooooooiiiiiiiiiinnnnnnnnnttttttttssssssss

Lisa [10]

Answer:

tttttttttttthankkssss!!!!!

3 0

2 years ago

Read 2 more answers

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

posledela

The question is incomplete, here is the complete question:

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

$2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag$

Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.

Answer: The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

Explanation:

We are given:

Moles of silver nitrate = 6 moles

For the given chemical reaction:

$2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag$

For copper metal:

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of copper metal

So, 6 moles of silver nitrate will react with = $\frac{1}{2}\times 6=3mol$ of copper metal

Moles of copper reacted = 3 moles

For copper(II) nitrate:

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 1 mole of copper(II) nitrate

So, 6 moles of silver nitrate will produce = $\frac{1}{2}\times 6=3mol$ of copper(II) nitrate

Moles of copper(II) nitrate produced = 3 moles

For silver metal:

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6 moles of silver nitrate will produce = $\frac{2}{2}\times 6=6mol$ of silver metal

Moles of silver metal produced = 3 moles

Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

6 0

3 years ago

What chemical reactions occur in a Nelson's cell?

Cerrena [4.2K]

Explanation:

The graphite anodes are suspended into the brine. During electrolysis, Cl ions are oxidized at the anode and chlorine gas goes out of the cell, while sodium ions are reduced at the mercury cathode forming sodium amalgam. ... Hydrogen gas is obtained as a by–product at the cathode.

8 0

1 year ago

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

Gre4nikov [31]

Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: Calculate moles of H₂O;

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

Also,

According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: Calculate Mass of O₂ and H₂ as;

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0

3 years ago