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3 years ago

In the Ideal Gas Law lab, how is the temperature of the hydrogen gas determined?

1 answer:

5 0

In the ideal gas law, the temperature of the hydrogen gas determined if the volume and pressure of the gas is known.

The ideal gas law states that the pressure of ideal gas is directly proportional to the absolute temperature of the gas.

PV = nRT

where;

P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
T is the temperature of the gas

The temperature of the gas is determined as follows;

$T = \frac{PV}{nR}$

Thus, we can conclude that in the ideal gas law, the temperature of the hydrogen gas determined if the volume and pressure of the gas is known.

Learn more here:brainly.com/question/24542515

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at a given tempature gas with a pressure of 150 kpa has volume of 0.8 l if the pressure decreases to 75kpa and the temperature r

vova2212 [387]

P*V/T=constant
so P*V= constant*T
if T doesn't change then
P*V= constant
so 150kPa*0.8L=75kPa*xL
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8 0

3 years ago

What is the molality of a solution if 100.0 g of glucose (C&Hi20e) were dissolved into 750. mL of water?

hodyreva [135]

Answer: The molality of solution is 0.740 m.

Explanation:

To calculate the mass of solvent (water), we use the equation:

$Density=\frac{Mass}{Volume}$

Volume of water = 750 mL

Density of water = 1 g/mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{750mL}\\\\\text{Mass of water}=750g$

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(C_6H_{12}O_6)$ = 100.0 g

$M_{solute}$ = Molar mass of solute $(C_6H_{12}O_6)$ = 180 g/mol

$W_{solvent}$ = Mass of solvent (water) = 750 g

Putting values in above equation, we get:

$\text{Molality of }C_6H_{12}O_6=\frac{100\times 1000}{180\times 750}\\\\\text{Molality of }C_6H_{12}O_6=0.740m$

Hence, the molality of solution is 0.740 m.

6 0

3 years ago

Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coeff

ahrayia [7]

Answer:

NH4Br + AgNO3 —> AgBr + NH4NO3

Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

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3 years ago