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3 years ago

Need help ASAP. Gave only 10 min

Chemistry

1 answer:

Kazeer [188]3 years ago

4 0

Answer:

Explanation:

If it interests you, you could check out Maxwell's demon. If he is not in operation (and he likely isn't), then the answer is D.

Heat always transfers from hot to cold -- at least at beginning levels of chemistry.

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1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t

slamgirl [31]

Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ? ; c₂ = 2.0 mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500 mL of water.

3 0

4 years ago

g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con

Tanzania [10]

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

$t_{1/2}=\frac{0.693}{k}}$ .....(1)

Rate law expression for first order kinetics is given by the equation:

$t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}$ ......(2)

where,

k = rate constant

$t_{1/2}$ =Half life of the reaction = $2.42\times 10^3 s$

t = time taken for decay process = ?

$[A_o]$ = initial amount of the reactant = 0.163 M

[A] = amount left after time t = 66.8% of $[A_o]$

[A]= $\frac{66.8}{100}\times 0.163 M=0.108884 M$

$k=\frac{0.693}{2.42\times 10^3 s}$

$t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}$

t = 1,409.19 s

1 minute = 60 sec

$t=\frac{1,409.19 }{60} min=23.49 min$

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

6 0

3 years ago

What is the ratio of surface area to volume for a sphere with the following measurements?

NeTakaya

Surface of a sphere=4πr²
volume of a sphere=(4/3)πr³
Data:
surface area =432 m²
volume=864 m³

ratio of surface area to volume=surface area / volume
ratio=432 / 864 =0.5

Answer. The ratio would be 0.5

8 0

4 years ago

Calculate the number of hydrogen atoms in a sample of hydrazine . Be sure your answer has a unit symbol if necessary, and round

gladu [14]

Answer:

$atoms \ H= 9.767x10^{24}atoms$

Explanation:

Hello!

In this case, considering that the mass of hydrazine is missing, we can assume it is 130.0 g (a problem found on ethernet). In such a way, since we need a mass-mole-atoms relationship by which we can compute moles of hydrazine given its molar mass (32.06 g/mol), then the moles of hydrogen considering one mole of hydrazine has four moles of hydrogen and one mole of hydrogen has 6.022x10²³ atoms (Avogadro's number); therefore, we proceed as shown below:

$atoms \ H=130.0gN_2H_4*\frac{1molN_2H_4}{32.06gN_2H_4} *\frac{4molH}{1molN_2H_4} *\frac{6.022x10^{23}atoms}{1molH}\\\\atoms \ H= 9.767x10^{24}atoms$

Notice 130.0 g has four significant figures, therefore the result is displayed with four as well.

Best regards!

7 0

3 years ago

Which of these particles are classified as hadrons? a. leptons c. mesons b. electrons d. neutrinos

disa [49]

The answer would be mesons.

Hope this image helps you! :)

4 0

4 years ago

Read 2 more answers