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3 years ago

What was the total number of calories lost when 10 grams of water at 80c is cooled to 60c?

Chemistry

1 answer:

pav-90 [236]3 years ago

3 0

Answer:

Q = -200 cal

Explanation:

Given data:

Calories lost = ?

Mass of water = 10 g

Initial temperature = 80°C

Final temperature = 60°C

Solution:

Specific heat capacity of water is 1 cal/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 60°C- 80°C

ΔT = -20°C

Q = m.c. ΔT

Q = 10 g. 1 cal/g.°C. -20°C

Q = -200 cal

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bearhunter [10]

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3 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°

miskamm [114]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H_1=157kJ$ ( × 4)

(2) $P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)$ $\Delta H_2=-1207kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]$

Putting values in above equation, we get:

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Hence, the $\Delta H^o_{rxn}$ for the reaction is -1835 kJ.

6 0

3 years ago

HURRY!!!! TIMED!!! WILL GIVE BRAINLIST!!!

aev [14]

Explanation:

(a) write a balanced equation for the reaction

CaCO3 + HCl --> CaCl2 + H2O + CO2

The balanced equation is given as;

CaCO3 + 2HCl → CaCl2 + H2O + CO2

(b) when the reaction was complete, 800 mL of carbon dioxide gas was collected. How many moles of calcium carbonate were used in the creation?

From the balanced reaction;

1 mol of CaCO3 reacts to produce 1 mol of CO2

1 mol of CO2 = 22.4 L of CO2

This means;

1 mol of CaCO3 reacts to produce 22.4L of CO2

x mol would produce 800ml (0.8 L) of CO2

1 = 22.4

x = 0.8

x = 0.8 * 1 / 22.4 = 0.0357 mol

Mass = Number of moles * Molar mass

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(d) If there was another contaminant in the sample that was not un reactive, would this have caused the percent yield of carbon dioxide to be higher, lower, or the same, explain your answer.

The same

An un reactive contaminant in the sample is most likely a catalyst. Catalysts only affect the rate of reaction. They do not affect yields of products.

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3 years ago

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earnstyle [38]

Answer:

Explanation:

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4 years ago

Will mark as Brainliest.

Likurg_2 [28]

D i took the test Enjoy :))))

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