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3 years ago

If 75 kJ of heat is transferred to 1200 g liquid water at 36°C, what would the

Chemistry

1 answer:

Aloiza [94]3 years ago

7 0

Answer:

50.9° (corrected to 3 significant figures)

Explanation:

Energy(J) = mass(kg) x specific heat capacity x temperature change

(E = mcΔT)

Let the new temperature be T.

75000 = (1.2 )(4186)(T-36)

14.930 = T -36

T = 50.9°

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At high pressures and moderate temperatures nitric oxide gas disproportionates rapidly according to the reaction 3NO(g) ⇌ N2O(g)

Pie

Answer:

3.8 x 10⁵

Explanation:

For the equilibrium : 3NO(g) ⇌ N2O(g) + NO2(g), the equilibrium constant in the terms of the concentrations of the gases in mol/L is

Kc = (NO) (N2O)/ (NO) ³ where (NO), (N2O) , (NO2) are the concentrations of the gases in mol/L . So

K= (x mol/ 1 L) (x mol/1L) / (7.5 x 10⁻⁶ mol /1 L) ³

x = mol of NO and NO2 at equilibrium

we have that

K = x²/ 7.5 x 10⁻⁶ = 1.9 x 10¹⁶

x = √ (7.5 x 10⁻⁶ x 1.9 x 10¹⁶) = 3.8 x 10⁵

∴ (N2O) = 3.8 x 10⁵

3 0

3 years ago

Read 2 more answers

Metals lose electrons and become

antoniya [11.8K]

Ionic compounds are molecules that form through the gain and loss of electrons. A metal atom that loses an electron takes on a positive electric charge .

6 0

3 years ago

Here’s another one guys please help.

ch4aika [34]

Answer:

Explanation:

i think B im not sure

3 0

3 years ago

What is the electrolysis of calcium hydroxide

spin [16.1K]

Electrolysis is the process of applying an electrical current to a solution in order to separate the components of that liquid phase solution on the basis of their charge. If we consider molten calcium hydroxide, the ions present will be:
Ca⁺²
OH⁻

The cations, the ones with a positive charge, will be reduced at the cathode. Therefore, calcium metal will be produced at the cathode. The anions will be oxidized at the anode, which means hydrogen gas will be produced at the anode.

4 0

3 years ago

Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is

Strike441 [17]

Answer:

For a: The molarity of commercial HCl solution is 12.39 M.

For b: The molality of commercial HCl solution is 16.79 m.

For c: The volume of commercial HCl solution needed is 2.42 L.

Explanation:

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

For a:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

For b:

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

For c:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

5 0

3 years ago