How much ice in grams at 0°C would have to melt to lower the temperature of 350 mL of water from 24°C to 6°C ? Assume the density of water is 1.0 g/mL.

Answers

Answer 1

In order to lower the temperature of 350 mL of water from 24°C to 6°C, we need to calculate the amount of heat energy that needs to be removed from the water. We can use the formula:

Q = mcΔT

where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of water is 4.184 J/g°C. Since the density of water is 1.0 g/mL, the mass of 350 mL of water is 350 g.

Q = (350 g) (4.184 J/g°C) (18°C)

Q = 30,276 J

To remove this much heat energy, we can use ice to cool the water. The heat of fusion of water is 333.55 J/g, which is the amount of heat energy required to melt one gram of ice at 0°C.

Q = (m)(333.55 J/g)

m = Q / 333.55 J/g

m = 30,276 J / 333.55 J/g

m = 90.4 g

So, 90.4 grams of ice at 0°C would have to melt to lower the temperature of 350 mL of water from 24°C to 6°C.


Related Questions

For the reaction 2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H₂O, how many grams of magnesium nitrate are produced from 9 grams of nitric acid, HNO3 ?​

Answers

This should be the answer


I’ll put the last part here


So, 10.52 grams of magnesium nitrate are produced from 9 grams of nitric acid, HNO3.

Which of the following indicates an exothermic reaction

Answers

Answer:

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings.”

a solution containing 1.00 g of an unknown nonelectrolyte liquid and 9.00 g of water has a freezing point of -3.33 *C. Calculate the molar mass of the unknown liquid in g/mol.

Answers

The depression in freezing point of water here is 3.33 °C. Then the molality of the solution is 1.79. From number of moles of solute is calculated as 0.0161. Then , the molar mass of the solute is 62.11 g.

What is depression in freezing point ?

The addition of a non-volatile salt into a solvent will decrease the freezing point of the solution than the pure solvent. The change in freezing point is proportional to the molality of the solution.

ΔTf = Kf m

Kf is called molal depression constant.

Kf for water = 1.86 °C Kg/mol

given ΔTf = 3.33 (0 - (-3.33))

then m = ΔTf /Kf =  3.33/ 1.86 °C Kg/mol = 1.79

Now, mass of water= 9 g = 0.009 kg

then number of moles = molality × mass of solvent

n = 0.009 kg × 1.79 = 0.0161.

mass of solute taken = 1 g

then, molar mass = 1/0.0161 = 62.11 g.

Therefore, the molar mass of the solute is 62.11 g.

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A bone taken from a garbage pile buried under a hill-side had 14C/12C ratio 0.477 times the ratio in a living plant or animal. What was the date when the animal was buried?​

Answers

Answer:

The date when the animal was buried can be estimated by using the decay rate of Carbon-14. The half-life of Carbon-14 is 5,730 years, so the date when the animal was buried can be estimated by taking the half-life and multiplying it by the logarithm of the ratio of the 14C/12C ratio of the bone and the 14C/12C ratio of a living plant or animal. In this case, 5,730 years x log(0.477) = 45,906 years ago.

Explanation:

Students in Ms. Brown’s class are on a field trip at the zoo. They observe the following biotic and abiotic factors in the zoo:

Rocks
Water
Trees
Grass
Monkeys
Birds

Ms. Brown would now like them to make a list of all the factors in the zoo ecosystem. Which of the items should be included on their list?

trees and grass
monkeys and birds
rocks, water, and trees
All abiotic and biotic factors of the zoo are part of the ecosystem.

Answers

Answer:

All abiotic and biotic factors of the zoo are part of the ecosystem.

Explanation:

An ecosystem is composed of all the living and non-living things in a particular area, and how they interact with one another. All the items listed, including rocks, water, trees, grass, monkeys, and birds, are part of the zoo ecosystem as they interact with each other and the environment.

For the following reactions, calculate the ionization constant of each reaction. CH3COOH (aq) + H2O (l) equation H3O+ (aq) + CH3COO-(aq); [CH3COOH] = 0.8M, [H3O+] = 7.6M, [CH3COO-] = 16.4MKa =

Answers

Ionization constant of a reaction is the ratio of the molar concentration of product to that of reactants. For the given  reaction, the ionization constant is 155.8 .

What is ionization constant ?

A compound or element gets ionized to form its anion or cation by the gain or lose of electrons.

Here, acetic acid, CH₃COOH ionizes to form hydronium ion and acetate ion. The ratio of product of molar concentrations of products to the product of molar concentrations of the reactants is called the reaction constant.

The ionization constant of the given reaction is written as:

K = [H₃O]⁺ [CH₃COO-]/[CH₃COOH]

Given, [CH₃COOH] = 0.8 M

[CH₃COO-] = 16.4 M

and  [H₃O]⁺ = 7.6 M.

then, K = 7.6 M × 16.4 M/ 0.8 M = 155.8

Therefore, the ionization constant of the reaction is 155.8.

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Read the information in the table below:



Which type of reaction is represented by B? (5 points)
Endothermic reaction
Exothermic reaction
Reaction between liquids
Reaction between solids

Answers

Answer:

it's exothermic reaction, becasue it have negative energy and the temperature increases

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