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3 years ago

A 2.6 liter container of nitrogen had a pressure of 3.2 atm. What volume will the gas occupy at a pressure of

Chemistry

1 answer:

Dominik [7]3 years ago

8 0

Answer:

A sample of oxygen gas occupies a volume of 250. ... volume will it occupy at 800. torr pressure? ... A 2.0 liter container of nitrogen had a pressure of 3.2 atm. ... A sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm producing

Explanation:

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Which natural law do biogeochemical cycles address?

emmainna [20.7K]

Im not completely sure but It might be Cycling of Matter

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3 years ago

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A gas mixture contains 3.00 atm of H2 and 1.00 atm of O2 in a 1.00 L vessel at 400K. If the mixture burns to form water while th

sleet_krkn [62]

Answer:

$p_{H_2O}=2.00atm$

Explanation:

Hello!

In this case, according to the following chemical reaction:

$2H_2+O_2\rightarrow 2H_2O$

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

$n_{H_2}=\frac{3.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0914molH_2 \\\\n_{O_2}=\frac{1.00atm*1.00L}{0.08206\frac{atm*L}{mol*K}*400K}=0.0305molH_2$

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

$n_{H_2O}^{by\ H_2} = 0.0914molH_2*\frac{2molH_2O}{2molH_2} =0.0914molH_2O\\\\n_{H_2O}^{by\ O_2} = 0.0305molO_2*\frac{2molH_2O}{1molO_2} =0.0609molH_2O$

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

$p_{H_2O}=\frac{0.0609molH_2O*0.08206\frac{atm*L}{mol*K}*400K}{1.00L}\\\\ p_{H_2O}=2.00atm$

Best regards!

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3 years ago

A b c d e f g h i j k l m n o p q r s t u v w x y z

enyata [817]

It’s B. Substitution hope this helps

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3 years ago

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Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined

Sunny_sXe [5.5K]

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here, suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.

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3 years ago

A substance undergoes a change. Which of the following indicates that the change was a chemical change?

hoa [83]

The answer is A. <span>The substance changed shape.

There you go!</span>

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3 years ago