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IRISSAK [1]
3 years ago
11

A plastic cup is filled with 15.5 grams of water. The cup is sealed and placed in the freezer. The water freezes and turns to ic

e. The next day, the ice is taken out of the cup and massed. What is the mass of the ice?
Chemistry
1 answer:
HACTEHA [7]3 years ago
7 0
It should be the same

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-Dominant- [34]
I think it’s B not rlly sure
5 0
2 years ago
A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t
andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

              = 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0
3 years ago
How many grams Glycylglycine (molecular mass 132.119 grams/mole) are required to make 1 liter at 0.5 Molar concentration. If you
SOVA2 [1]

Explanation:

The given data is as follows.

        Concentration of solution = 0.5 M

        Volume of solution = 1 L

Molar mass of Glycylglycine = 132.119 g/mol

As molarity is the number of moles present in liter of solvent.

Mathematically,      Molarity = \frac{\text{No. of moles}}{\text{Volume}}

Hence, calculate the number of moles as follows.

            No. of moles = Molarity × Volume

                                  = 0.5 M \times 1 L

                                  = 0.5 mol

Therefore, mass of glycylglycine = mol × molar mass

                                                      = 0.5 mol \times 132.119 g/mol

                                                      = 66.06 g

Thus, we can conclude that 66.06 g glycylglycine is required.

8 0
3 years ago
A radioactive element reduces to 5.00% of its initial mass in
Stolb23 [73]

The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months

To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Amount remaining (N) = 5%

Original amount (N₀) = 100%

<h3>Number of half-lives (n) =?</h3>

N₀ × 2ⁿ = N  

5 × 2ⁿ = 100

2ⁿ = 100/5

2ⁿ = 20

Take the log of both side

Log 2ⁿ = log 20

nlog 2 = log 20

Divide both side by log 2

n = log 20 / log 2

<h3>n = 4.32</h3>

Thus, 4.32 half-lives gas elapsed.

Finally, we shall determine the half-life of the element. This can be obtained as follow.

Number of half-lives (n) = 4.32

Time (t) = 500 years

<h3>Half-life (t½) =? </h3>

t½ = t / n

t½ = 500 / 4.32

t½ = 115.74 years

Multiply by 12 to express in months

t½ = 115.74 × 12

<h3>t½ ≈ 1389 months </h3>

Therefore, the half-life of the radioactive element in months is approximately 1389 months

Learn more: brainly.com/question/24868345

8 0
3 years ago
All of the following are balanced equations. State the
Advocard [28]

Answer: a: reactants Na-2 Cl-2

Products: Na-2 Cl-2;

b: reactants P-1 Cl-13 H-6 Products P-1 H-6 Cl-13

c: reactants P-4 H-12 O-16

Products H-12 P-4 O-16

Explanation: since these equations are balanced the atoms on of element on the reactants side will be same as the atoms of the same element of the product side

7 0
3 years ago
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