All categories

3 years ago

Why use fire in the industrial revolution? HELP PLEASE❤️❤️❤️

Chemistry

1 answer:

REY [17]3 years ago

7 0

Answer:

Some of the most important American inventions during the Industrial Revolution included the telegraph, the sewing machine, telephone, cotton gin, the practical light bulb, and vulcanized rubber. Manchester, England was the center of the textile industry during the Industrial Revolution.

Explanation:

You might be interested in

What is the Ka of a 0.0796 M solution of nitrous acid (HNO2) with a pH of 2.95?

Vadim26 [7]

Answer:

Coefficient = 1.58

Exponent = - 5

Explanation:

pH = 2.95

Molar concentration = 0.0796M

Ka = [H+]^2 / [HA]

Ka = [H+]^2 / 0.0796

Therefore ;

[H+] = 10^-2.95

[H+] = 0.0011220 = 1.122 × 10^-3

Ka = [H+] / molar concentration

Ka = [1.122 × 10^-3]^2 / 0.0796

Ka = (1.258884 × 10^-6) / 0.0796

Ka = 15.815 × 10^-6

Ka = 1.58 × 10^-5

Coefficient = 1.58

Exponent = - 5

4 0

4 years ago

What is the acceleration of the object’s motion? 0.5 m/s2 -0.5 m/s2 2 m/s2 -2 m/s2

djverab [1.8K]

Answer: -2m/s2

Explanation:

Using the following equation ; acceleration = Change in velocity / time

i.e a = v - u / t

where 'a' = acceleration

v = final velocity

u = initial velocity

t = time

Therefore; from the graph we have acceleration to be, 0 - 6m/s / 3s = -2m/s2

5 0

3 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas