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3 years ago

7

Calculate the mass of water formed from the complete combustion of 5.10 g of methane, ch4, in the reaction ch4 (g) + 2 o2 (g) --

> co2 (g) + 2 h2o (g).

1 answer:

Anton [14]3 years ago

4 0

CH₄ + 2O₂ ---> CO₂ + 2H₂O
16g..................................18g×2

16g CH₄ --- 36g H₂O
5,1g CH₄ --- X
X = (5,1×36)/16
X = 11,475g H₂O

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How many micrograms (ug) are in 3.4 x 10^-5 ounces (oz)

marysya [2.9K]

Answer:

964ug

Explanation:

The problem here involves converting from one unit to another.

We are to convert from ounces to micrograms.

1ug = 1 x 10⁻⁶g

1oz = 28.35g

So we first convert to grams from oz then take to ug:

Solving:

1oz = 28.35g

3.4 x 10⁻⁵oz will then give 3.4 x 10⁻⁵ x 28.35 = 9.64 x 10⁻⁴g

So;

1 x 10⁻⁶g = 1ug

9.64 x 10⁻⁴g will give $\frac{9.64 x 10^{-4} }{1 x 10^{-6} }$ = 9.64 x 10²ug or 964ug

8 0

2 years ago

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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