Which type of circuit has a string of lights and one light goes out but the others stay lit

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq) , as desc

vlada-n [284]

Answer:

0.87g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

MnO2(s) + 4HCl(aq) —> MnCl2(aq) + 2H2O(l) + Cl2(g)

Step 2:

Data obtained from the question. This includes the following:

Volume (V) of Cl2 obtained = 235mL

Temperature (T) = 25°C

Pressure (P) = 805 Torr

Step 3:

Conversion to appropriate unit.

For Volume:

1000mL = 1L

Therefore, 235mL = 235/1000 = 0.235L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 25°C

Temperature (Kelvin) = 25°C + 273 = 298K

For Pressure:

760 Torr = 1 atm

Therefore, 805 Torr = 805/760 = 1.06 atm

Step 4:

Determination of the number of mole of Cl2 produced. This is illustrated below:

The number of mole (n) of Cl2 produced can be obtained by using the ideal gas equation as follow:

PV = nRT

Volume (V) = 0.235L

Temperature (T) = 298k

Pressure (P) = 1.06 atm

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n)

PV = nRT

Divide both side by RT

n = PV /RT

n = (1.06 x 0.235)/(0.082 x 298)

n = 0.01 mole

Therefore 0.01 mole of Cl2 is produced from the reaction.

Step 5:

Determination of the number of mole MnO2 that produce 0.01 mole of Cl2. This is illustrated below:

MnO2(s) + 4HCl(aq) —> MnCl2(aq) + 2H2O(l) + Cl2(g)

From the balanced equation above,

1 mole of MnO2 produced 1 mole of Cl2.

Therefore, it will take 0.01 mole to MnO2 to also produce 0.01 mole of Cl2.

Step 6:

Converting 0.01 mole of MnO2 to grams.

This is illustrated below:

Number of mole MnO2 = 0.01 mole

Molar Mass of MnO2 = 55 + (2x16) = 87g/mol

Mass of MnO2 =?

Mass = number of mole x molar Mass

Mass of MnO2 = 0.01 x 87

Mass of MnO2 = 0.87g

Therefore, 0.87g of MnO2 is needed for the reaction.

8 0

3 years ago

When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac

STatiana [176]

Answer:

$\Delta _{comb}H=-2,093\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

$Q_{rxn}+Q_{cal}=0$

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

$Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ$

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

$n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}$

Best regards!

4 0

3 years ago

A typical refrigerator is kept at 4˚C, and a soda can has a pressure of

ss7ja [257]

Answer: The new pressure will be 1.42 atm

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=1.18atm\\T_1=4^0C=(4+273)K=277K\\P_2=?\\T_2=60^0C=(60+273)K=333K$

Putting values in above equation, we get:

$\frac{1.18}{277}=\frac{P_2}{333}\\\\P_2=1.42$

Hence, the new pressure will be 1.42 atm

7 0

3 years ago

What is the molarity of a solution of Cu2SO4 if 250.0 mL of solution contains 14.1 g of Cu2SO4?

balu736 [363]

Copper(I) compounds in aqueous solutions are unstable and disproportionate:
Cu₂SO₄ = Cu + CuSO₄
Necessary for dissolution of non-aqueous solvent.

M(Cu₂SO₄)=223.16 g/mol
m(Cu₂SO₄)=14.1 g
v=0.250 L

n(Cu₂SO₄)=m(Cu₂SO₄)/M(Cu₂SO₄)

c=n(Cu₂SO₄)/v=m(Cu₂SO₄)/(vM(Cu₂SO₄))

c=14.1/(0.250*223.16)=0.253 mol/L

0.253 M

5 0

3 years ago

Read 2 more answers

Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz

stiv31 [10]

Answer: Freezing point of a solution will be $-1.16^0C$