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Marta_Voda [28]
3 years ago
13

A particular brand of gasoline has a density of 0.737 g/mL at 25 ∘C. How many grams of this gasoline would fill a 15.7 gal tank

(1US gal=3.78L)?
Chemistry
1 answer:
sergiy2304 [10]3 years ago
7 0
Th above problem asked us to find the amount of gasoline in grams given a density of 0.737 g/mL at 25°C and a volume of 15.7 gal.

To solve this problem, we must use the formula of density which is mass over volume.

D=\frac{mass}{volume}

Base on the formula, we can find the mass of gasoline in by multiplying both sides by the volume. That is,
mass=D(Volume)

Note that we must cancel out the unit with a remaining g as the unit of mass. We must convert galloons to liters to mL.. That is

1 gal=3.78L and 1L=1000mL
15.7 gal x\frac{3.78 L}{1 gal} x\frac{1000 mL}{1 L}=59346 mL

So,
mass=(0.737 g/mL x 59346 mL)=43738 g

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A 1.775g sample mixture of potassium hydrogen carbonate is decomposed by heating. if the mass loss is 0.275g what is the percent
Marina86 [1]

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

<h3>What is a decomposition reaction?</h3>

A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.

  • Step 1: Write the balanced equation for the decomposition of KHCO₃.

2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)

The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.

  • Step 2: Calculate the mass of KHCO₃ that formed 0.275 g of CO₂.

In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.

0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃

  • Step 3: Calculate the mass percentage of KHCO₃ in the sample.

There are 1.25 g of KHCO₃ in the 1.775 g sample.

%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%

A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.

Learn more about decomposition reactions here: brainly.com/question/14219426

7 0
2 years ago
Which process is an example of a chemical change?
WARRIOR [948]

Answer:

A is the closest thing. You change the composition of the steak. You don't in any of the others.

Explanation:

Usually when you cook something, you are doing something to the composition of the object being cooked. A steak might not be obvious, but boiling an egg should be.

Chopping a tree is something physical. You are removing mass in such a way that the tree will fall. There's nothing chemical about that.

Heating a cup of tea looks like it might be chemical. After all steam is sometimes given off which looks like it is chemical. It's not. The water in the tea is just changing phase.

Drying clothes in a dryer. Again, this looks like something might have changed. After all the mass of the clothes just became less. But all you are doing is separating two masses (leaving one of them behind).

8 0
3 years ago
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Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to
posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

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

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

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5 0
3 years ago
What is the final volume?
zaharov [31]

Answer:

Option A. 9.4 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 8 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 343 K

Final volume (V₂) =?

V₁ / T₁ = V₂ / T₂

8 / 293 = V₂ / 343

Cross multiply

293 × V₂ = 8 × 343

293 × V₂ = 2744

Divide both side by 293

V₂ = 2744 / 293

V₂ = 9.4 L

Therefore, the final volume of the gas is 9.4 L

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3 years ago
Atoms like sodium and lithium often lose electrons, which makes them
zvonat [6]
More reactive than others
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