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Dmitrij [34]
3 years ago
9

If you collect 1.75 L of hydrogen gas during a lab experiment when the room temperature is 23oC and the barometric pressure is 1

05 kPa how many moles of hydrogen will you have
Chemistry
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

n=0.0747mol

Explanation:

Hello,

In this case, since we can consider hydrogen gas as an ideal gas, we check the volume-pressure-temperature-mole relationship by using the ideal gas equation:

PV=nRT

Whereas we are asked to compute the moles given the temperature in Kelvins, thr pressure in atm and volume in L as shown below:

n=\frac{105kPa*\frac{0.009869atm}{1kPa}*1.75L}{0.082\frac{atm*L}{mol*K}*(23+273.15)K} \\\\n=0.0747mol

Best regards.

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Iron that is tin-plated does not rust. Why not?
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Rust is iron oxide, the corrosion product of iron when exposed to the oxygen in the air. Tin is not iron, so you cannot produce iron oxide from the corrosion of tin. Because the layer of tin on the surface of the steel prevents atmospheric oxygen and moisture from contacting the steel.

8 0
3 years ago
Read 2 more answers
At 500K, the equilibrium constant for the reaction N2O4D2NO2 is 1.5 x 10^3. What is the equilibrium constant for the reaction: 6
Veronika [31]

Answer:

K = 2.96x10⁻¹⁰

Explanation:

Based on the initial reaction:

N2O4 ⇄ 2NO2; K = 1.5x10³

Using Hess's law, we can multiply this reaction changing K:

3 times this reaction:

3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹

The inverse reaction has a K of:

6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;

<h3>K = 2.96x10⁻¹⁰</h3>

3 0
3 years ago
5. __NH3 + __O2 &gt;&gt;&gt;&gt;&gt;__ NO +__ H20
inn [45]
2 NH3+ 2 O2 —> 2 NO+ 3 H2O
5 0
3 years ago
R-COOH compounds behave as _____. <br> A) bases <br> B) acids <br> C) esters <br> D) ketones
lakkis [162]
<span>R-COOH 
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6 0
3 years ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
Karo-lina-s [1.5K]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

8 0
3 years ago
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