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3 years ago

How much energy, in kJ, is released when 27.0 grams of water is cooled from 112C to -15C?

Chemistry

1 answer:

Lady_Fox [76]3 years ago

6 0

Answer: Q

−

796.6

Explanation is right here instead

The equation for heat is

(

−

)

Q = heat in Joules

m = mass in grams

= final temp

= initial temp

= Specific Heat Capacity

For this problem

275

38.4

85.2

4.18

275

(

38.4

−

85.2

)

4.18

275

(

−

46.8

)

4.18

−

796.6

The negative Joule value means heat is being lost in an exothermic reaction.

Explanation:

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Fault-block mountains are formed when a shift takes place along a fault line and

worty [1.4K]

One side drops into fault

7 0

3 years ago

El agua salubre es aquella que tiene más sales disueltas que el agua dulce. En el análisis de una muestra de aguas se encontró q

Slav-nsk [51]

Answer:

0.39 % m/m; 0.42 % m/v; 0.18 % v/v; 4200 ppm; 0.044 mol·L⁻¹; 0.041 mol/kg;

0.089 equiv/L; 0.000 74; 0.999 26

Explanation

Data:

Mass of MgCl₂ = 3.8 g

Volume of solution = 900 mL

Density of solution = 1.09 g/mL

Density of MgCl₂ = 2.32 g/cm³

Calculations

1. Percent m/m

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times 100 \, \%\\\\\text{Total mass} = \text{900 mL} \times \dfrac{\text{1.09 g}}{\text{1 mL}} = \text{981 g}\\\\\text{Mass percent} = \dfrac{\text{3.8 g}}{\text{981 g}} \times 100 \,\% = \textbf{0.39 \% m/m}$

2. Percent m/v

$\text{Mass-by-volume percent} = \dfrac{\text{Mass of component}}{\text{Total volume}} \times 100 \, \%\\\\\text{ Mass-by-volume percent } = \dfrac{\text{3.8 g}}{\text{900 mL}} \times 100 \,\% = \textbf{0.42 \% m/v}$

3. Percent v/v

$\text{Volume percent} = \dfrac{\text{Volume of component}}{\text{Total volume}} \times 100 \, \%\\\\\text{Volume of MgCl}_{2} = \text{3.8 g} \times \dfrac{\text{1 mL}}{\text{2.32 g}} = \text{1.64 mL}\\\\\text{ Mass-by-volume percent } = \dfrac{\text{1.64 mL}}{\text{900 mL}} \times 100 \,\% = \textbf{0.18 \% v/v}$

4. Parts per million

$\text{Ppm} = \dfrac{\text{milligrams of solute}}{\text{litres of solution}} = \dfrac{\text{3800 mg}}{\text{0.900 L}} = \textbf{ 4200 ppm}$

5. Molar concentration

$\text{Molar concentration} = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\\text{Moles of MgCl}_{2} = \text{3.8 g} \times \dfrac{\text{1 mol}}{\text{95.21 g}} = \text{0.040 mol}\\\\\text{Molar concentration} = \dfrac{\text{0.040 mol}}{\text{0.900 L}} = \textbf{0.044 mol/L}$

6. Molal concentration

$\text{Molal concentration} = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}$

Mass of water = Mass of solution - mass of solute = 981 g - 3.8 g = 977 g = 0.977 kg

$\text{Molar concentration} = \dfrac{\text{0.040 mol}}{\text{0.977 kg}} = \textbf{0.041 mol/kg}$

7. Normality

The normality is the number of equivalents per litre of solution.

The normality of an ion equals the molar concentration times the charge on the ion.

Thus, the normality of MgCl₂ is twice the molar concentration.

Normality = 2 × 0.044 mol·L⁻¹ = 0.089 equiv·L⁻¹

8. Mole fraction of solute

$\chi_{\text{solute}} = \dfrac{n_{\text{solute}}}{n_{\text{total}}}$

Moles of MgCl₂ = 0.040 mol

$\text{Moles of water} = \text{977 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{54.23 mol}$

Total moles = n₂ + n₁ = 0.040 mol + 54.23 mol = 54.27 mol

$\chi_{2} = \dfrac{\text{0.040 mol}}{\text{54.23 mol}} = \mathbf{0.00074}$

9. Mole fraction of solvent

χ₁ = 1 - χ₂ = 1 - 0.000 74 = 0.999 26

4 0

3 years ago

Nitrogen dioxide reacts with carbon monoxide to produce nitrogen monoxide and carbon dioxide. NO2(g) + CO(g) ⟶ NO(g) + CO2(g) A

Naya [18.7K]

Answer: The rate law for the reaction is $\text{Rate}=k[NO_3][CO]$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)$

The intermediate reaction of the mechanism follows:

Step 1: $2NO_2(g)\rightleftharpoons NO_3(g)+NO(g);\text{ (fast)}$

Step 2: $NO_3(g)+CO(g)\rightarrow NO_2(g)+CO_2(g);\text{(slow)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[NO_3][CO]$

Hence, the rate law for the reaction is written above.

5 0

4 years ago

A new process converts potential energy of an object to kinetic energy. Neither work nor eat is added to the system. No reaction

blagie [28]

Answer:

Change in height of the object

Explanation:

Change in height of the object

Since ,

We know potential energy can be written as mgh where

P.E. = mgh

M = mass ,

g = gravity constant

h = height

Kinetic energy can be written as 0.5 mv²

So , the Potential energy = kinetic energy

Then ,

velocity of object = √2 gh

Since g is a constant ,

Hence ,

Height of object will determine the velocity.

5 0

4 years ago

Liquids have a density greater than 1 g/m? Sometimes true? True? never true?

kirza4 [7]

Answer:

Water has a density of 1 g/m so it could be sometimes true

Explanation:

4 0

4 years ago

Read 2 more answers