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3 years ago

If the reactants have a potential energy of 10.2 kJ/mol and the products have a potential

Chemistry

1 answer:

Citrus2011 [14]3 years ago

4 0

Answer:

The ΔH is 5.5 kJ/mol and the reaction is endothermic.

Explanation:

To calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them:

Combustion enthalpy = ΔH = ∑H products - ∑Hreactants

In this case:

ΔH = 15.7 kJ/mol - 10.2 kJ/mol= 5.5 kJ/mol

An endothermic reaction is one whose enthalpy value is positive, that is, the system absorbs heat from the environment (ΔH> 0).

<u><em>The ΔH is 5.5 kJ/mol and the reaction is endothermic.</em></u>

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What change in volume results if 60 mL of a gas is cooled from 33 C to 5 C?

Eva8 [605]

Answer:

Change in volume on changing temperature from 33 $^{\circ}C$ to 5 $^{\circ}C$ is 5.49 mL

Explanation:

Initial volume of gas = V = 60 mL

Assuming final volume of gas to be V' mL

Initial temperature = T = 33 $^{\circ}C$ = 306 K

Final temperature = T' = 5 $^{\circ}C$ = 278 K

The relationship between volume and temperature of gas at constant pressure is shown below

$\displaystyle \frac{V}{V'}=\displaystyle \frac{T}{T'} \\\displaystyle \frac{60\textrm{ mL}}{V'} = \displaystyle \frac{306\textrm{ K}}{T} \\V' = 54.51 \textrm{ mL} \\\textrm{Change in volume} = \left ( V-V' \right ) \\\textrm{Change in volume} = \left ( 60-54.51 \right )\textrm{ mL} \\\textrm{Change in volume} = 5.49 \textrm{ mL}$

Change in volume on changing temperature = 5.49 mL

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3 years ago

A man carrying a torch walks though a jungle on a very dark night. Suddenly, he sees two eyes looking down from a tree that seem

Brut [27]

D) The eyes in the tree are reflecting light from the torch into the man's eyes.

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3 years ago

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

How many grams of chromium metal are plated out when a constant current of 8.00 A is passed through an aqueous solution containi

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Answer:

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Explanation:

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3 years ago

Is frictional force desirable?

gavmur [86]

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3 years ago