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Allisa [31]
3 years ago
5

Is a new can of paint open or closed system

Chemistry
1 answer:
Brums [2.3K]3 years ago
8 0

Answer:

I think it might be a closed system

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What is the molarity of a solution in which 0.321 g of calcium chloride is dissolved in 1.45 L water?
lana66690 [7]

Answer:

Molarity = 0.002 M

Explanation:

Given data:

Mass of calcium chloride = 0.321 g

Volume of water = 1.45 L

Molarity of solution = ?

Solution:

Molarity = number of moles / volume in litter.

We will calculate the number of moles of calcium chloride first.

Number of moles = mass/molar mass

Number of moles = 0.321 g/ 110.98 g/mol

Number of moles = 0.003 mol

Molarity:

Molarity = 0.003 mol / 1.45 L

Molarity = 0.002 M

7 0
2 years ago
The swimmer moves her hand down and to the left and her body goes forward
Ganezh [65]

Answer:Newton’s first law

Explanation:

4 0
2 years ago
46. Which of the following is not a form of matter?
Mice21 [21]
C cuz it just doesn’t make sense
3 0
3 years ago
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How many cu atoms are present in a piece of sterlingsilver jewelry weighing 33.24 g ? (sterling silver is a silver–copper alloy
emmasim [6.3K]

The mass of piece of sterling silver jewelry is 33.24 g. It contains 92.5% silver Ag by mass. Since, sterling silver is an alloy of Ag-Cu thus, percentage of Cu will be:

% Cu=100-92.5=7.5%

Thus, mass of copper will be:

m_{Cu}=\frac{7.5}{100}\times 33.24 g=2.493 g

Molar mass of Cu is 63.546 g/mol, thus, number of moles of Cu can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Putting the values,

n=\frac{2.493 g}{63.546 g/mol}=0.03923 mol

Now, in 1 mole of Cu there are 6.303\times 10^{23}atoms thus, in 0.03923 mol, number of Cu atoms will be:

0.03923 mol\times 6.303\times 10^{23}atoms=2.363\times 10^{22} atoms

Thus, number of atoms of Cu will be2.363\times 10^{22} atoms.


8 0
3 years ago
Please help!!!!!! BRAINLIEST to right answer! :))
notsponge [240]

Explanation:

d. endothermic change as

heat is added to solid ice to change it to liquid water

6 0
3 years ago
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