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ozzi
3 years ago
7

Balance the equation ___ CaCO3 -> ____ CaO + ____ CO2

Chemistry
1 answer:
Degger [83]3 years ago
3 0

Answer:

CaCO3 -> CaO + CO2

Explanation:

Woahhhh, did you balance it yourself just then?

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How many moles of calcium (Ca) atoms are there in 77.4 g of Ca?
Vinvika [58]
Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
8 0
3 years ago
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Which statement describes the bonds in iron sulfate, FeSO4?
Feliz [49]

___

Regarding the bonds in FesO₄, Fe and S have an ionic bond, while S and O have covalent bonds.

Elements form bonds to increase their stability. The main types of bonds are:

  • Metallic bonds: they are formed between metals and the electrons are in a delocalized cloud.
  • Ionic bonds: they are formed between metals (lose electrons) and nonmetals (gain electrons)
  • Covalent bonds: they are formed between nonmetals, which share electrons.

Regarding the bonds in FesO₄:

  • Fe is a metal and S a nonmetal, thus they will form ionic bonds.
  • S and O are both nonmetals, thus they will form covalent bonds.

Regarding the bonds in FesO₄, Fe and S have an ionic bond, while S and O have covalent bonds.

Learn more: brainly.com/question/23882847

5 0
2 years ago
1. Drug dependence from drug abuse​
tia_tia [17]

Answer:

✅

Explanation:

DON'T USE DRUGS BECAUSE IT IS GOING DEATH

4 0
3 years ago
Which definition best describes a titration?
Juliette [100K]

Answer:

the answer is C

Explanation:

4 0
3 years ago
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The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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