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barxatty [35]
3 years ago
13

What mass to the nearest tenth of a gram, is needed to balance this chemical

Chemistry
1 answer:
Andrej [43]3 years ago
6 0

none of the above. is the answer

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For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0
3 years ago
What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are
damaskus [11]

Answer:

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

7 0
3 years ago
Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s
liubo4ka [24]

Answer:

the heavy one

Explanation:

the heavy one because heavy things and break things and the light one can't

7 0
3 years ago
Please help, and answer as much as you can. ​
n200080 [17]

Answer:

∴ Fractional distillation is the technique used to separate the fraction

Explanation:

A: Refinery gas

B: Gasoline fraction

C: Naphtha

D: Kerosene

E: Diesel Oil

F: Fuel oil fraction

G: Lubricating fraction

H: Bitumen

5 0
2 years ago
Which of the following have the most similar properties?
rjkz [21]

Answer:

             Option-B (Carbon and Silicon)

Explanation:

                  Among the given pairs only carbon and silicon have the most similar properties. This is because,

Sodium and Magnesium belong to different groups. Sodium present in Group I has one electron in its valence shell and capable of transferring only one electron while, Magnesium present in Group II have two electrons in its valence shell and is capable of donating two electrons. Hence, both show different properties.

Example:

                                           2 Na  +  Cl₂    →     NaCl

                                           Mg  +  Cl₂    →    MgCl₂

As shown in reactions when Sodium and Magnesium are treated with Cl₂ they give a products with different proportions.

Carbon and Silicon show almost same properties because both belong to Group IV hence both are capable of forming four bonds. Also, they share the same property of self linkage in making a long chains.

Argon and Chlorine also belong to two different groups. Argon is present in Group VIII (Noble Gases) and Chlorine is present in Group VII (Halogens). Hence, Argon is an inert specie which is non reactive while Chlorine gives different reaction easily.

Potassium and Calcium belong to different groups. Potassium present in Group I has one electron in its valence shell and capable of transferring only one electron while, Calcium present in Group II have two electrons in its valence shell and is capable of donating two electrons. Hence, both show different properties.

Example:

                                           2 K  +  Cl₂    →     KCl

                                           Ca  +  Cl₂    →    CaCl₂

As shown in reactions when Potassium and Calcium are treated with Cl₂ they give a products with different proportions.

6 0
4 years ago
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