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3 years ago

The relationships among which variable quantities are expressed by the ideal gas law equation?

Chemistry

2 answers:

Eva8 [605]3 years ago

8 0

The variable quantities are expressed by the ideal gas law equation are; 

pressure, volume, temperature, number of moles

This question is simply based on defining the ideal gas law.

Now, A gas is considered to ideal if its particles are so far from each other in such a manner that they don't exhibit any forces of attraction between themselves. Now, in real life this is not possible but under high temperatures and pressure, we can have something close to it and that's why ideal gas laws are very important.

This law states that states that the pressure, temperature, number of moles and volume of a gas are related to each other by the formula;

PV = nRT

Where;

P is pressure

V is volume

n is number of moles

T is temperature

R is ideal gas constant (This is fixed and not variable)

The variable quantities are expressed by the ideal gas law equation are;

pressure, volume, temperature, number of moles

Read more at; brainly.in/question/5212853

iris [78.8K]3 years ago

4 0

Answer:

I am positive it is: pressure, volume, temperature, number of moles.

Explanation:

Here is the equation for the ideal gas law: PV=nRT

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Cuántos electrones de Valencia tiene Ga

Zinaida [17]

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

Primero realizas la configuración electrónica que es la que te puse allá arriba.

Después miras el nivel en que termina, puede ser 1, 2, 3, 4 etc.

Entonces como el último número de la configuración electrónica es 4, entonces ese es el nivel

Y los electrones de el último nivel son los de Valencia

4s2, 4p1 sumas 2+1 que son los electrones que se encuentran en el último nivel.

por eso hay 3 electrones de valencia.

4 0

3 years ago

What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?

Brut [27]

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

7 0

3 years ago

Determine whether each description represents a genotype or a phenotype.

ANEK [815]

Answer:

phenotype,phenotype,genotype,genotype,

genotype

Explanation:

phenotype is physical appearance and genotype is just like

yy Tt

8 0

3 years ago

Read 2 more answers

When equal volumes of 0.5 M HCl and 0.5 M Ca(OH)2 are mixed, the resulting solution is

Mademuasel [1]

The concentration of mixed solution = 0.5 M

<h3> Further explanation </h3>

Given

0.5 M HCl

0.5 M Ca(OH)₂

Required

The concentration

Solution

Molarity from 2 solutions :

Vm Mm = V₁. M₁ + V₂. M₂

m = mixed solution

V = volume

M = molarity

V = mixed volume

1 = solution 1

2 = solution 2

Vm = V₁+V₂

Equal volumes⇒V₁=V₂, and Vm = 2V, then equation becomes :

2V.Mm = V(M₁+M₂)

2V.Mm = V(0.5+0.5)

Mm=0.5 M

8 0

3 years ago

2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

8 0

2 years ago