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strojnjashka [21]
2 years ago
10

Which of the following is a class 3 surgical candidate

Chemistry
1 answer:
nataly862011 [7]2 years ago
5 0
Ikr behehbenekebe sgwhebejebeb
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A particular car has a gas mileage of 24.5 miles per gallon. If the cost of gas is $4.25 per gallon, and the car travels at a co
tester [92]

Answer:

$30.39

Explanation:

alot of math

3 0
2 years ago
Jeff has 10 grams of water and 10 grams of vegetable oil in separate containers. Both liquids have a temperature of 24°C. Jeff h
Anna11 [10]
I think the answer would be that vegetable oil is a better conductor of heat than water.
8 0
2 years ago
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(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
2 years ago
The information below describes a redox reaction.
Alchen [17]

Answer:

3

Explanation:

You have to mutiply the silver reaction by 3 in order to substract the electrons

3 0
2 years ago
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A 15.8 g sample contains 3.60 g F, 4.90 g H, and 7.30 g C. What is the percent composition of hydrogen in this sample?
deff fn [24]

Answer:

The correct answer is "32%".

Explanation:

The given values:

Weight of H,

= 4.9 g

Weight of sample,

= 15.8 g

Now,

The weight percentage of C will be:

= \frac{Weight \ of \ C}{Total \ weight}\times 100

By substituting the values, we get

= \frac{4.9}{15.8}\times 100

= 32 \ percent

3 0
2 years ago
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