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Korvikt [17]
3 years ago
15

How many calories are required to raise the temperature of 105 g of water from 30.0°c to 70.0°c?

Chemistry
1 answer:
ioda3 years ago
5 0
<span>4.200 calories. The formula is rise in heat (H) = the specific heat capacity (c) * rise in heat in degrees Fahrenheit (T) and volume of liquid (m) . H= cTm. Water, the typical comparison, has a specific heat capacity of 1. So to raise the Fahrenheit temperature of 105 grams of water 40 degrees, you multiply 105 * 40 * 1 = 4,200.</span>
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If the number is not a proper coefficient, how would you make it one?
noname [10]
I’m pretty sure you add, multiply, subtract, & divide
3 0
2 years ago
You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution
vfiekz [6]

Answer:

450. g of 0.173 % KCN solution contains 779 mg of KCN.

Explanation:

Mass of the solution = m

Mass of the KCN in solution = 779 mg

Mass by mass percentage of KCN solution = 0.173%

(m/m)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

0.173\%=\frac{779 mg}{m}\times 100

m=\frac{779 mg}{0.173}\times 100= 450,289 mg

1 mg = 0.001 g

m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g

450. g of 0.173 % KCN solution contains 779 mg of KCN.

6 0
3 years ago
Which observation does not indicate that a chemical reaction has occurred?
castortr0y [4]

Answer:

change in the total mass of substances

6 0
3 years ago
How many moles are in 29.5 grams of Ax?
JulsSmile [24]

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

Ar = 39.95 g/mole

The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

= \frac{29.5}{39.95} \\\\= 0.74 \ mole

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.

<em>"Your question seems to be missing the correct symbol for the element" </em>

Argon = Ar

Learn more here:brainly.com/question/4628363

6 0
2 years ago
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7 0
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