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3 years ago

How many moles of methanol are required to produce 4.19 moles of water

Chemistry

1 answer:

Harrizon [31]3 years ago

6 0

Explanation:

every two moles of methanol will need three moles of oxygen gas and products 2 miles of carbon dioxide and 4 miles of water

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What would cause the pressure of a gas to increase?

IrinaK [193]

Answer: The pressure of a gas will increase when there is a decrease in the volume of the gas.

Explanation: according to Boyle's law, the volume of a gas will decrease when the pressure is increased at constant temperature and vice versa.

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3 years ago

What causes water to increase in density?

murzikaleks [220]

Explanation:

temperature changes affect seawater density as water cools its density increases. As water cools H2O molecules pack more closely together because the molecules are vibrating less at low temperatures and take up less volume. The same number of water molecules in smaller volume results higher density

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3 years ago

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If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

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

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e

Sunny_sXe [5.5K]

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

$N_{2}O_{4} \rightarrow 2NO_{2}$

Let us assume that x concentration of $N_{2}O_{4}$ is present at the initial stage. Therefore, according to the ICE table,

$N_{2}O_{4} \rightarrow 2NO_{2}$

Initial : x 0

Change : - 0.1 $2 \times 0.1$

Equilibrium : (x - 0.1) 0.2

Now, expression for $K_{p}$ of this reaction is as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

Putting the given values into the above formula as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

$2 = \frac{(0.2)^{2}}{(x - 0.1)}$

$2 \times (x - 0.1) = (0.2)^{2}$

x = 0.12

This means that $P_{N_{2}O_{4}}$ = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

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3 years ago

Which of the following is an example of a physical change?(1 point)

user100 [1]

Dissolving sugar in water

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3 years ago

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