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Fudgin [204]
2 years ago
7

2. The three sub-atomic particles that make up the atom are

Chemistry
1 answer:
AlladinOne [14]2 years ago
5 0

Answer:

Protons

Neutrons

Electrons

Explanation:

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10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of
Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and 7.2m^3 respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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

where,

P_1 = initial pressure of gas = 10 kPa

P_2 = final pressure of gas = 15 kPa

V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 50^oC=273+50=323K

T_2 = final temperature of gas = 75^oC=273+75=348K

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

\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}

V_2=7.2m^3

The new volume of carbon dioxide gas is 7.2m^3

Now we have to calculate the new density of carbon dioxide gas.

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

Formula for new density will be:

\rho_2=\frac{P_2M}{RT_2}

where,

P_2 = new pressure of gas = 15 kPa

T_2 = new temperature of gas = 75^oC=273+75=348K

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

\rho = new density

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

\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

\rho_2=0.2281g/L

The new density of carbon dioxide gas is 0.2281 g/L

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How does kinetic energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at
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The larger the kinetic energy of the vehicle, the larger the amount of energy will be needed to stop the vehicle, meaning that faster vehicles have a larger stopping distance
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Los metales son buenos conductores de la electricidad?
stealth61 [152]
English please i cant understand
8 0
2 years ago
Copper(II) sulfate pentahydrate, CuSO4 ·5 H2O, (molar mass 250 g/mol) can be dehydrated by repeated heating in a crucible. Which
prohojiy [21]

Answer:

The water lost is 36% of the total mass of the hydrate

Explanation:

<u>Step 1:</u> Data given

Molar mass of CuSO4*5H2O = 250 g/mol

Molar mass of CuSO4 = 160 g/mol

<u>Step 2:</u> Calculate mass of water lost

Mass of water lost = 250 - 160 = 90 grams

<u>Step 3:</u> Calculate % water

% water = (mass water / total mass of hydrate)*100 %

% water = (90 grams / 250 grams )*100% = 36 %

We can control this by the following equation

The hydrate has 5 moles of H2O

5*18. = 90 grams

(90/250)*100% = 36%

(160/250)*100% = 64 %

The water lost is 36% of the total mass of the hydrate

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2 years ago
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kotykmax [81]
Fe^2 O^3 + 6HCl --> 2FeCl^3 + 3H^2 O
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3 years ago
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