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Gre4nikov [31]
3 years ago
13

True or false: The product of a chemical reaction is always different than the original reactants.

Chemistry
1 answer:
TEA [102]3 years ago
5 0

Answer:

The answer is true

Explanation:

Products in chemical reactions are rearranged during the reaction, the atoms end up in different combinations in the products. This makes the product new substances that are chemically different than the reactants

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Which law states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constan
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<span>Boyles law states that the volume of a gas is proportional to the moles of the gas when pressure and temperature are kept constant.    </span>
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Menstruation occurs on a ______ interval from puberty until menopause
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A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 209.kPa. T
disa [49]

Answer : The final pressure of the gas will be, 26.8 kPa

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.

P\propto \frac{1}{V}

or,

PV=k

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of the gas = 209 kPa

P_2 = final pressure of the gas = ?

V_1 = initial volume of the gas = 10.0 L

V_2 = final volume of the gas = 78.0 L

Now put all the given values in this formula, we get the final pressure of the gas.

209kPa\times 10.0L=P_2\times 78.0L

P_2=26.8kPa

Therefore, the final pressure of the gas will be, 26.8 kPa

3 0
3 years ago
A steel container with a movable piston contains 2.00 g of helium which was held at a constant temperature of 25 °C. Additional
shtirl [24]

Answer: D) 1.00 g

Explanation:

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = 2.00 L

V_2 = final volume of gas = 3.00 L

n_1 = initial moles of gas  =\frac{\text {Given mass of helium}}{\text {molar mass of helium}}=\frac{2.00g}{4g/mol}=0.500mol

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{2.00L}{3.00L}=\frac{0.500mol}{n_2}

n_2=0.75mole

Mass of helium =moles\times {\text {molar mass}}=0.75\times 4=3.00g

Thus mass of helium added = (3.00-2.00) g = 1.00 g

4 0
3 years ago
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