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2 years ago

A fish tank holds 31 gal of water. How many kilograms of water are in the fish tank?

Chemistry

1 answer:

Oksanka [162]2 years ago

8 0

Answer:

3.785411784

Explanation:1 gallon (gal) = 3.785411784 kilogram (kg). Gallon (gal) is a unit of Volume used in Standard system. Kilogram (kg) is a unit of Weight used in Metric system. Please note this is volume to weight conversion, this conversion is valid only for pure water at temperature 4 °C.

US oz = 28.349523125 g

US fl oz = 29.5735295625 ml (milliliters) = 29.5735295625 g (grams) of pure water at temperature 4 °C.

US cup = 8 fl oz = 236.5882365 ml (milliliters) = 236.5882365 g (grams) of pure water at temperature 4 °C.

The US gallon is equal to 3.785411784 liters and defined as 231 cubic inches. Kilograms also can be marked as kilogrammes in UK.

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3 years ago

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3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

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3 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

musickatia [10]

Answer:

$T_2=335.42K=62.27^oC$

Explanation:

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In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

$T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC$

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Leto [7]

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