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3 years ago

Which information is sufficient to differentiate a

Chemistry

2 answers:

blagie [28]3 years ago

8 0

Answer: (3) The reactivity of each sample with water

Explanation: Since sodium dissolves in water, you could place the two elements in water and see which of the two dissolve

Novay_Z [31]3 years ago

3 0

"The reactivity of each sample with water" is sufficient to differentiate the sample of sodium from a sample of silver.

Explanation :

Silver does not undergo any reaction with water. Also, silver is insoluble in water under normal conditions. Thus it remains in its metallic state when it comes in contact with water.

On the other hand, sodium is highly reactive in water. It forms a colorless solution in water, which is basic. This basic solution is sodium hydroxide and hydrogen gas. The reaction of sodium and water is exothermic.

$2 N a(s)+2 H_{2} O \rightarrow 2 N a O H(a q)+H_{2}(g)$

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An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):

$V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol$

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

$n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol$

$V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21$

$P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm$

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.

3 0

3 years ago

What properties of matter are intensive independent of their size

Paha777 [63]

Answer:

temperature, refractive index, density, and hardness of an object

Explanation:

it is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system

8 0

3 years ago

What are the missing coefficients for the skeleton equation below? Cr(s) + Fe(NO3)2(aq) → Fe(s) + Cr(NO3)3(aq)

postnew [5]

Answer:

2 3 3 2

Explanation:

7 0

3 years ago

The atomic number is the number of protons in the atom of the element

ipn [44]

Answer:

True

Explanation:

If you look closley at the nucleus, you don't count the neutrons just the prtons which then effect the electrons.

Good luck :)

5 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago