Ask question

Ask question

All categories

3 years ago

8

Write the correct name for each compound. a. N2O5 b. Si2F4 c. S3F d. OF2 e. H4P6 f. C2O4 g. HF3

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Answer:

a. Boron trifluoride

b. Propane

c. Dinitrogen pentoxide

d. Carbon Dioxide

e. Silicon Octafluroride?

Explanation:

Glad to help :)

You might be interested in

When oil and water are mixed vigorously, the oil is broken up into tiny droplets that are dispersed throughout the water. this d

Bumek [7]

Emulsion or heterogeneous mixture

4 0

2 years ago

If you are using a thermometer you are measuring the

Slav-nsk [51]

Temperature of somewhere.

6 0

3 years ago

For which compound does 0.256 mole weigh 12.8 g?

mylen [45]

Number of moles = $\frac{mass (g)}{Molar mass}$
so to calculate molar mass (Molecular weight of compound):
Molecular weight = $\frac{mass (g)}{number of moles}$
= $\frac{12.8 g}{0.256 mole}$ = 50 g / mole
Now we calculate molecular weight of each compound in choices:
a) C₂H₄O = 44
b) CO₂ = 44
c) CH₃Cl = 50.4 ALMOST 50 so this is the correct answer
d) C₂H₆ = 30

6 0

2 years ago

Acetone has a density of 0.7857 g/cm^3. What is the volume in milliliters of 7.70g of acetone

Brums [2.3K]

Pretty sure this is it

8 0

3 years ago

Read 2 more answers

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago