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Alex Ar [27]
3 years ago
10

If temperature is held constant for an ideal gas, as P and V change for a given gas sample, the PV of products will_____. increa

se decrease be constant change
Chemistry
2 answers:
noname [10]3 years ago
6 0

Answer:

The PV of products will be constant.

Explanation:

Ideal gas equation-   PV = nRT

where P is pressure of gas, V is volume of gas, n is number of moles of gas, R is gas constant and T is temperature (in Kelvin) of gas.

Now, number of moles of a gas remain constant with change in state variables.

So, if T is constant then the term "nRT" remins constant.

As PV is equal to nRT therefore PV will be constant if temperature is held constant.

In another way we can say if P is increased then V will decrease or vice versa.

vodka [1.7K]3 years ago
3 0

Answer:

PV=nRT

Explanation:

V=<u>R</u><u>T</u><u>n</u>

P

rearrangement gives

  • PV=nRT
  • R=<u>P</u><u>V</u>

nT

where P=pressure

V=volume

n=number of moles

R=ideal gas(0.0820atmdm/3 mol/k)

T=temperature in kelvin

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Which statements would be expected to be true for Organesson? Select all that apply.
NISA [10]

Answer:It has the lowest effective nuclear charge of the four new elements.

5 0
2 years ago
112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0
2 years ago
Explain the differences between bronze and brass, and steel and cast iron. Which is best used for cooking wares?
Dennis_Churaev [7]

Answer:

Steel and cast iron

Explanation:

They are all metal but assuming that you are finding the best material for your pan i suggest going for steel or cast iron

8 0
3 years ago
1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
Greeley [361]

Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

7 0
3 years ago
A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the
Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

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

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

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

\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}

P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

4 0
2 years ago
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