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irina1246 [14]
2 years ago
11

4. If 100.0 g of water is cooled from 25.3°C to 16.9°C, what quantity of

Chemistry
1 answer:
Allisa [31]2 years ago
3 0

Answer:

Dived the numbers by each other

Explanation:

25.3/19.9

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Please help asap,
Finger [1]
10. Capital C and D represent products of chemical reaction, the capital A and B represent reactants, <span>the lower case letter represent coefficients (how many atoms or molecules in chemical reaction).
12. According to </span><span>Le </span>Chatelier's principle (if<span> the concentration is changed, that will shift the equilibrium to the side that would reduce that change in concentration)</span> <span>the equilibrium shift to the left.
13. </span>According to Le Chatelier's principle the equilibrium shift to the right.
14. According to Le Chatelier's principle (<span>When the reaction is </span>exothermic<span>, heat is included as a product)</span> the equilibrium shift to the right.
5 0
3 years ago
An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of
kherson [118]

Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

MAss glucose = moles glucose* molecular weight of glucose

MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

We need 8.11 grams of glucose for this solution

6 0
3 years ago
Commercial grade HCl solutions are typically 39.0% (by mass) HCl in water. Determine the molality of the HCl, if the solution ha
marta [7]

Molality of the solution is defined as the number of moles of a substance dissolved divided by the mass of the solvent:

Molality = number of moles / solvent mass

From the concentration of 39% (by mass) of HCl in water, we construct the following  reasoning:

in 100 g solution we have 39 g hydrochloric acid (HCl)

number of moles  = mass / molecular weight

number of moles of HCl = 39 / 36.5 = 1.07 moles

solvent (water) mass = solution mass - hydrochloric acid mass

solvent (water) mass = 100 - 39 = 61 g

Now we can determine the molality:

molality = 1.07 moles / 61 g = 0.018

8 0
3 years ago
How do i solve it and what is the answer? ​
Masja [62]

Umm there’s no picture

6 0
3 years ago
What would you need to do to calculate the molality of 10 g of NaCl in 2 kg of
jasenka [17]

Answer:

O B. Convert the 10 g of NaCl to moles of NaCl.

Explanation:

The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.

(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl

58.44 is the molar mass of NaCl

m=0.1711 mol NaCl/2 kg H2O

m=0.085557837

7 0
3 years ago
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