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3 years ago

Plz help I have to finish test in 1 hour!!!!

Chemistry

1 answer:

AnnZ [28]3 years ago

6 0

Answer:

I think its the first one. unless there is more to the question

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An ideal gas described by Ti=291K, Pi=1.50bar, and Vi=13.3L is heated at constant volume until P=15.0bar. It then undergoes a re

Salsk061 [2.6K]

Answer:

$W=-4601.4J$

Explanation:

Hello,

In this case, the steps are:

$291K,1.50bar, 13.3L \rightarrow 15.0bar 13.3L,T_2\rightarrow T_2, V_2, 1.50bar\rightarrow 291K,1.50bar, 13.3L$

In such a way, the work per mole (w) for that isothermal process turns out:

$w=RTln(\frac{P_1}{P_2} )=8.314\frac{J}{mol*K}*291K*\frac{1.50bar}{15.0bar} \\\\w=-5570.8\frac{J}{mol}$

In addition, if the moles are required, since it is an ideal gas:

$n=\frac{PV}{RT}=\frac{1.50bar*13.3L}{0.083\frac{bar*L}{mol*K}*291K} =0.826mol$

So the work is:

$W=-5570.8\frac{J}{mol} *0.826mol=-4601.4J$

Best regards.

8 0

3 years ago

You have a vial that may contain lead(II) cations or barium cations, but not both. Available to you are solutions of ammonium su

Bess [88]

Answer:

We should add, the KBr solution.

Explanation:

The vial contains Ba²⁺ and Pb²⁺

Our solutions are: (NH₄)₂ SO₄, NaF and KBr

We dissociate the solutions:

(NH₄)₂ SO₄ → 2NH₄⁺ + SO₄⁻²

NaF → Na⁺ + F⁻

KBr → K⁺ + Br-

When we add ammonium sulfate and sodium fluorine, we are adding sulfates and fluorines, but these two anions make precipitate both cations, we can not determine, if the vial contains Ba²⁺ or Pb²⁺

SO₄⁻²(aq) + Ba²⁺ (aq) → BaSO₄ (s) ↓

SO₄⁻² (aq)+ Pb²⁺(aq) → PbSO₄ (s) ↓

2F⁻ (aq) + Ba²⁺ (aq) → BaF₂ (s) ↓

2F⁻ (aq) + Pb²⁺ (aq) → PbF₂ (s) ↓

When we add the KBr, bromides only make precipitate with Pb²⁺, so the vial has only Pb²⁺ cation:

2Br⁻ (aq) + Ba²⁺ (aq) → BaBr₂ (aq)

2Br⁻ (aq) + Pb²⁺ (aq) → PbBr₂ (s) ↓

3 0

3 years ago

Determine whether each statement is a description of a physical property or a chemical property.

Sunny_sXe [5.5K]

Answer:

Paper=chemical

Salt=physical

Explanation:

5 0

3 years ago

What is meant by the term specific heat?

Alex73 [517]

Answer:

the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

8 0

3 years ago

There are 43.2 g of carbon to 14.4 g of hydrogen in a sample of methane (CH4). What percent of 37.8 g of methane is carbon?

EastWind [94]

Using the law of constant proportions which says that within the same compound, elements exist in fixed ratios.

Therefore; we can use the ratio of total mass to the mass of carbon, to determine the amount of carbon in another sample.

Mass C / Mass CH4 = Mass C / Mass CH4

43.2 g / 57.6 g = Mass C / 37.8 g

Mass C = 37.8 g × 43.2 g / 57.6 g

= 28.35 g

Hence; the percentage of carbon will be;

=(28.35/ 37.8 )× 100%

= 75 %

Thus; 75% of 37.8 g of methane is carbon

8 0

4 years ago