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olga_2 [115]
3 years ago
10

Calculate the amount of heat needed to convert 190.0 g of ice at -15 °C to water at 35 °C. Cp of ice=2.09J/g*c

Chemistry
1 answer:
salantis [7]3 years ago
6 0

Answer:

The amount of heat required to melt 25 grams of ice is 8,350 Joules or 2,000 calories.

Explanation:

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What’s the SI unit of the amount of isopropyl alcohol in a beaker?
Illusion [34]

Answer:

mole

Explanation:

The mole in chemistry is used to represent the amount of any substance. Just like quantifying everyday things like a dozen, score, gross etc, it is a convenient unit of quantity of particles. A mole denotes 6.02 x 10²³particles of a susbstance.

Therefore, a mole is the standard unit(SI) for the amount of isopropyl alcohol in a beaker.

7 0
2 years ago
Which method would be best to separate a mixture of oil and water?
rosijanka [135]

Answer:

distillation ​

Explanation:

Actually, distillation is used to separate liquids from nonvolatile solids, as in the separation of alcoholic liquors from fermented materials, or in the separation of two or more liquids having different boiling points, as in the separation of gasoline, kerosene, and lubricating oil from crude oil.

7 0
2 years ago
Which equation has x = 5 as the solution?
Natali5045456 [20]

Answer:

x-10 should be the answer

8 0
2 years ago
19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?
Alecsey [184]

Answer:

m_{H_2O}=3.384gH_2O

Explanation:

Hello,

In this case, the chemical reaction is:

Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O

In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:

m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O

Regards.

4 0
2 years ago
Some oxygen gas takes up a volume of 2.00 liters at 0.99 atm and 273 K. Its volume doubles and its temperature decreases to 137
nataly862011 [7]

Answer:

The answer to your question is : letter B. 0.25 atm

Explanation:

To solve this problem we need to use the combined gas law:

    <u>P₁V₁</u>   =    <u>P₂V₂</u>

      T₁              T₂

Data

P1 = 0.99 atm  V1 = 2 l  T1 = 273K

P2 = ?               V2 = 4 l   T2 = 137K

Now, the clear P2 from the equation and we get

       P2 = P1V1T2 / T1V2

Substitution         P2 = (2 x 0.99 x 137)/(273 x 4)

       P2 = 271.26 / 1092

Result       P2 = 0.248 atm ≈ 0.25 atm

5 0
3 years ago
Read 2 more answers
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