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kozerog [31]
3 years ago
14

Will mark as Brainliest.

Chemistry
1 answer:
Bess [88]3 years ago
7 0
The answer should be A, increase the concentration of both
You might be interested in
A piston chamber filled with ideal gas is kept in a constant-temperature bath at 25.0°C. The piston expands from 25.0 mL to 75.0
larisa [96]

Answer : The work done by the system is, 2.2722 J

Explanation :

The expression used for work done in reversible isothermal expansion will be,

w=nRT\ln (\frac{V_2}{V_1})

where,

w = work done = ?

n = number of moles of gas  = 0.00100 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 25^oC=273+25=298K

V_1 = initial volume of gas  = 25 mL

V_2 = final volume of gas  = 75 mL

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

w=0.00100mole\times 8.314J/moleK\times 298K\times \ln (\frac{75}{25})

w=2.722J

Therefore, the work done by the system is, 2.2722 J

8 0
3 years ago
Part A
Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

  • <u>7.33 × 10 ⁻³ c</u>

Explanation:

<u>1) Data:</u>

a) m = 9.11 × 10⁻³¹ kg

b) λ =  3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

<u>2) Formula:</u>

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

Solve for s:

  • s = h / (m.λ)

Substitute:

  • s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg ×  3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c =  3.00 10⁸ m/s

  • s =  2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

  • 2.06 × 10 ⁻³⁴ m.

Explanation:

<u>1) Data:</u>

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ =  ?

<u>2) Formula:</u>

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

  • λ = h / (m.s)

  • h is Planck's constant: h= 6.626×10⁻³⁴J.s

<u>3) Solution:</u>

  • λ = h / (m.s)

Substitute:

  • λ =  6.626×10⁻³⁴J.s / ( 0.0459 kg ×  70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.  

5 0
3 years ago
A hydrocarbon contains 85.7% carbon and the remainder
Viefleur [7K]

Answer:

CH₂ ;  67.1 %

Explanation:

To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation

Assume 100 grams of the compound.

# mol C = 85.7 g / 12.01 g/mol = 7.14 mol

# mol H = 14.3 g /  1.008 g/mol = 14.19 mol

The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C

So the empirical formula is CH₂

For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄  reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.

We need to calculate the moles of  NaBH₄ ( M.W = 37.83 g/mol )

1.203 g  NaBH₄ / 37.83 g/mol =  0.0318 mol

Theoretical yield from balanced chemical equation:

0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆

Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol =  0.440 g

% yield = 0.295 g/ 0.440 g x 100 = 67.1 %

6 0
3 years ago
What structural units make up metallic solids?
Anettt [7]

Answer:

metal atom

Explanation:

metallic solid has layers makes it soft and it also has electron that can carry charge and conduct electricity.

4 0
3 years ago
2) Calculate the percent composition of each element in Mgso,
iogann1982 [59]

Answer:

2)

\% Mg=20.2\%\\\\\% S=26.6\%\\\\\% O=53.2\%

3)

\% Ag=93.1\%\\\\\% O=6.9\%

Explanation:

Hello!

2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:

\% Mg=\frac{24.305}{120.36 } *100\%=20.2\%\\\\\% S=\frac{32.06}{120.36 } *100\%=26.6\%\\\\\% O=\frac{64.00}{120.36 } *100\%=53.2\%

3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:

\% Ag=\frac{215.74}{231.74} *100\%=93.1\%\\\\\% O=\frac{16.00}{231.74} *100\%=6.9\%

Best regards!

7 0
3 years ago
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