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3 years ago

It has more protons than Cl but less than K

Chemistry

1 answer:

timama [110]3 years ago

8 0

Answer: Argon (Ar), which has 18 protons.

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Which statement correctly describes a characteristic that a scientific measuring toll should have?

Gwar [14]

Answer:

Explanation:

To be accurate, it must be able to make measurements that are close to the actual value.

6 0

3 years ago

Given the reaction at equilibrium:

VLD [36.1K]

Answer:

decrease

Explanation:

because the equilibrium will shift to right side

8 0

3 years ago

Some hypothetical alloy is composed of 12. 5 wt% of metal a and 87. 5 wt% of metal b. If the densities of metals a and b are 4.

gavmur [86]

Answer:

The number of

atoms in the unit cell is 2

Explanation:

7 0

2 years ago

A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely

Rzqust [24]

Answer:

1.59 atm

Explanation:

The reaction is:

$N_{2}O_{3}(g) - - -> NO_{2}(g)+NO(g)$

The dalton's law tell us that the total pressure of a mixture of gases is the sum of the partial pressure of every gas.

So after the reaction the total pressure is:

$P_{total}=P_{NO_{2}}+P_{NO}$

we don't include $N_{2}O_{3}$ because it decomposed completely.

Assuming ideal gases

PV=nRT

P= pressure, V= volume of the container, n= mol of gas, R=constant of gases and T=temperature.

so moles of $N_{2}O_{3}$ is:

$n_{N_{2}O_{3}}=\frac{P_{1}V}{RT_{1}}$

from the reaction stoichiometry (1:1) we have that after the reaction the number of moles of each product is the same number of moles of $N_{2}O_{3}$ .

$n_{NO_{2}}=\frac{P_{1}V}{RT_{1}}$

$n_{NO}=\frac{P_{1}V}{RT_{1}}$

The partial pressure of each gas is:

$P_{NO_{2}}=\frac{n_{NO_{2}*R*T_{2}}}{V}$

$P_{NO}=\frac{n_{NO}*R*T_{2}}{V}$

so total pressure is:

$P_{total}=(n_{NO_{2}}+n_{NO})*\frac{R*T_{2}}{V}$

replacing the moles we get:

$P_{total}=(2*\frac{P_{1}V}{RT_{1}})*\frac{R*T_{2}}{V}$

We know that T2=3*T1

replacing this value in the equation we get:

$P_{total}=2*\frac{P_{1}V}{RT_{1}}*\frac{R*3T_{1}}{V}$

$P_{total}=6*P_{1} = 6*0.265 atm = 1.59 atm$

5 0

3 years ago

A balloon originally had a volume of 4.39 L at 44C and a pressure of 729 torr to what temperature must the balloon be cooled

VMariaS [17]

The new temperature : 11.56 °C

<h3>Further explanation </h3>

Boyle's law and Gay Lussac's law

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

P1 = initial gas pressure (N/m² or Pa)

V1 = initial gas volume (m³)

P2 = final gas pressure

V2 = final gas volume

T1 = initial gas temperature (K)

T2 = final gas temperature

V₁=4.39 L

T₁=44+273=317 K

P₁ = 729 torr = 0,959211 atm

V₂=3.78 L

P₂= 1 atm

$\tt \dfrac{0.959211\times 4.39}{317}=\dfrac{1\times 3.78}{T_2}\\\\T_2=\dfrac{1\times 3.78\times 317}{0.959211\times 4.39}\\\\T_2=284.559~K=11.56~C$

3 0

3 years ago