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allsm [11]
3 years ago
15

What is the formula for calcium carbonate?

Chemistry
1 answer:
mote1985 [20]3 years ago
6 0
The formula is CaCO3
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Will give brainlist need answers asap
lord [1]

Answer:

2nd and 4th ones

Explanation:

More sugar dissolves in water than salt

  more aspirin dissolves in water than CO2 ...but less aspirin than table salt (NaCl)

5 0
2 years ago
13. What is the charge of an aluminum ion that has 13 protons and 10 electrons?
Ksenya-84 [330]

Answer:

+ 3

Ex planation:

Al3+ indicates an ion of aluminum having a charge of + 3. I.e., since an aluminum atom normally has 13 protons and 13 electrons, this ion has 10 electrons (-10 charge) and 13 protons (+ 13 charge) giving it a charge of + 3 (-10 + 13 = +3).

hope this helps!

7 0
3 years ago
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A substance has twice the number of particles as 12 grams of carbon-12. How many
klasskru [66]

Answer:

D)2.0

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by N_0.

Also, it is the number of particles in exactly 12.000 g of isotope carbon 12.

Avogadro constant:-

N_a=6.023\times 10^{23}

Thus,

Number of particles as 12 grams of carbon-12 constitutes 1 mole

<u>Twice the number of particles as 12 grams of carbon-12 constitutes 1*2 mole</u>

Thus, moles in the substance = 2.0 moles

3 0
3 years ago
Liquid nitrogen, which has a boiling point of −195.79°C, is used as a coolant and as a preservative for biological tissues. Is t
Ray Of Light [21]

Answer:

Explanation:

Entropy is measure of disorder so as we lower the temperature of gas , its entropy decreases .

Hence at - 200°C entropy of nitrogen will be less than that at - 190°C .

At freezing point ,

entropy of fusion  = latent heat / freezing temperature

= .71 kJ / ( 273 - 210 )

= 710 / 63 J mol⁻¹ K⁻¹ .

= 11.27 J mol⁻¹ K⁻¹ .

entropy of fusion = 11.27 J mol⁻¹ K⁻¹ .

8 0
3 years ago
I have four questions.
navik [9.2K]
Abca I think it’s right not sure
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3 years ago
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