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Goshia [24]
3 years ago
6

Will give brainliest.

Chemistry
2 answers:
Umnica [9.8K]3 years ago
8 0
Pretty sure it’s 40 (:
Sergeeva-Olga [200]3 years ago
3 0
40. Because sodium chloride is NaCl
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frozen [14]
1.78L x (3.00M/1L) = 5.34
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3 years ago
Oxygen is an example of a(n)
stepladder [879]

Answer:

C. element. oxygen is an element

Explanation:

oxygen is element #8

7 0
3 years ago
If 6.02×1023 atoms of element Y have a mass of 28.09 g, what is the identity of Y?
Mazyrski [523]

Silicon is the element having a mass of 28.09 g

<u>Explanation</u>:

  • Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023 \times 10^23 atoms. One mole of each element can produce one mole of compound.
  • The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023 \times10^23).
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8 0
3 years ago
All of the following are forms of acid precipitation EXCEPT ________.
Aneli [31]
Dihydrogen oxide is the right answer. Dihydrogen oxide is just 2 hydrogen and 1 oxygen which is H2O or water.
6 0
3 years ago
A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0
3 years ago
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