PLEASE HELP WITH CHEMISTRY !! BRAINLIEST AND 10 POINTS !!

The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi

Setler79 [48]

Answer:

Explanation:

mass fraction N₂ : He : CH₄ : C₂H₆ : : 15 : 5 : 60 : 20

mole fraction N₂ : He : CH₄ : C₂H₆ : : 15/28 : 5/4 : 60/16 : 20/30

mole fraction N₂ : He : CH₄ : C₂H₆ : : .5357 : 1.25 : 3.75 : .67

Total mole fractions = .5357 + 1.25 + 3.75 + 0.67 = 6.2057

mole fraction of N₂ = .5357 / 6.2057 = .0877

mole fraction of He = 1.25 / 6.2057 = .20

mole fraction of CH₄ = 3.75 / 6.2057 = .6043

mole fraction of C₂H₆ = .67 / 6.2057 = .108

Partial pressure = total pressure x mole fraction

Partial pressure of N₂ = 1200 kPa x .0877 = 105.24 kPa

Partial pressure of He = 1200 kPa x .20 = 240 kPa

Partial pressure of CH₄ = 1200 kPa x .6043 = 725.16 kPa

Partial pressure of C₂H₆ = 1200 kPa x .108 = 129.6 kPa

6 0

3 years ago

Copper has a density of 8.920 g/cm3. What mass of copper will occupy 45cm3? What volume of copper will have a mass of 1.0 kg?

mixas84 [53]

Density = mass/volume
so rearranged mass = volume x density
mass = 8.920 x 45 = 401.4g

rearrange (there are 1000grams in 1kg)
volume = mass/density
volume = 1000/8.920
volume = 112.1076233cm3

5 0

4 years ago

Gold has a density of about 20g/mL. If I measured out 2mL of gold, I would expect it to have a mass of about ___ grams.

yan [13]

The answer is 40 grams. If gold is 20 grams per mL and you have 2 mLs, you get 40 grams of gold. Just add 20 + 20 or multiply 2 x 20 to get 40.

4 0

3 years ago

The ph of a 0.15 m aqueous solution of naz (the sodium salt of hz) is 10.7. what is the ka for hz?

andrew-mc [135]

Answer:
NaZ + H2O <----> HZ + OH-
[OH-] = [HZ]
[NaZ] = 0.15 M
pH = 10.7, so the pOH = 14-10.7 = 3.3; [OH-] = 10^-3.3 = 5.01E-4
kb = [OH-][HZ]/[NaZ] = (5.01E-4)^2/0.15 = 1.67E-6
ka = kw/kb = 1E-14/1.67E-6 = 5.99E-9 = 6.0E-9

5 0

3 years ago

Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) an

mafiozo [28]

Answer:

$\rm K$ : approximately $22.0\%$ .

$\rm Cl$ : approximately $19.9\%$ .

$\rm O$ : approximately $27.0\%$ .

$\rm Al$ : $31.0\%$ .

Explanation:

Consider a $100\; \rm g$ sample. There would be $69.0\% \times 100\; \rm g = 69.0\; \rm g$ of $\rm KClO_{3}$ and $100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g$ of $\rm Al$ .

Out of that $69.0\; \rm g$ of $\rm KClO_{3}$ :

$31.9\%$ (by mass) would be $\rm K$ : $31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g$ .

$28.9\%$ (by mass) would be $\rm Cl$ : $28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g$ .

$39.2\%$ (by mass) would be $\rm O$ : $39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g$ .

Overall, the composition (by mass) of each element in this mixture would be:

$\rm K$ :

$\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}$ .

$\rm Cl$ :

$\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}$ .

$\rm O$ :

$\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}$ .

$\rm Al$ :

$\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}$ .

7 0

3 years ago