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seraphim [82]
3 years ago
5

Last friday the atmospheric pressure in our 2nd year lab was measured as 731 mmhg. Calculate the temperature at which water woul

d boil at this pressure.
Chemistry
1 answer:
faust18 [17]3 years ago
7 0

Answer:

p1/T1=p2/T2

760mmHg/212°F=731mmHg/T2

T2= 203.91°F

760mmHg/100°C=731mmHg/T2

T2= 96.18°C

Explanation:

You'd have to choose in which units you want to express the temperature.

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Which of the following is a physical property of a substance?
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B boiling point https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/03%3A_Matter_and_Energy/3.05%3A_Differences_in_Matter%3A_Physical_and_Chemical_Properties#Summary
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3 years ago
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd
kenny6666 [7]

Answer:

The answer is "= 0.078 \ kg \ H_2".

Explanation:

calculating the moles in CH_4 =\frac{PV}{RT}

                                                =\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol

Eqution:

CH_4 +H_2O \to  3H_2+ CO \ (g)

Calculating the amount of H_2 produced:

= 12.9 \ mol CH_4 \times  \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2

So, the amount of dihydrogen produced = 0.078 \frac{kg}{s}

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3 years ago
Speed of an electron to orbit radius wire linear charge density is _________
ch4aika [34]

Answer:

A long uniformly charged wire has charge density λ=0.16μλ=0.16μC/m. 

7 0
3 years ago
Consider the reaction below.
Law Incorporation [45]

Answer:

  • <u>First choice: 0.042</u>

Explanation:

Given decomposition reaction:

  • 1PCl₅ (g) ⇄ 1PCl₃ + 1Cl₂(g)

Equilibrium constant:

  • K_{eq}=\frac{[PCl_3]^1[Cl_2]^1}{[PCl_5]^1}

Stoichiometric coefficients and powers equal to 1 are not usually shown as they are understood, but I included them in order to shwow you how they intervene in the equilibrium expressions: each concentration is raised to a power equal to the respective stoichiometric coefficient in the equilibrium equation.

So, your calculations are:

K_{eq}=\frac{(0.020M)(0.020M)}{0.0095M}=0.042M

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3 years ago
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