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jok3333 [9.3K]
2 years ago
14

Consider the balanced equation.

Chemistry
1 answer:
viktelen [127]2 years ago
8 0
28.4% is the answer and that is ur answer
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Mass x height x gravity is the formula to calculate:
Sindrei [870]

Answer:

Potential energy

Explanation:

p.e = mgh

8 0
3 years ago
I don't unerstand what I'm supposed to do for these
shepuryov [24]
You just need to multiply the terms and transform to kj by dividing by 1000
1) 10.4kj
2) 14.5 kj 
3) 44.8 kj
4) 1.04 kj

and there you go.
8 0
3 years ago
Read 2 more answers
What is the Ka of a 0.0796 M solution of nitrous acid (HNO2) with a pH of 2.95?
Vadim26 [7]

Answer:

Coefficient = 1.58

Exponent = - 5

Explanation:

pH = 2.95

Molar concentration = 0.0796M

Ka = [H+]^2 / [HA]

Ka = [H+]^2 / 0.0796

Therefore ;

[H+] = 10^-2.95

[H+] = 0.0011220 = 1.122 × 10^-3

Ka = [H+] / molar concentration

Ka = [1.122 × 10^-3]^2 / 0.0796

Ka = (1.258884 × 10^-6) / 0.0796

Ka = 15.815 × 10^-6

Ka = 1.58 × 10^-5

Coefficient = 1.58

Exponent = - 5

4 0
4 years ago
according to iupac nomeclature rules which set of chemical name and formula for the same compound is correct?
kumpel [21]

Answer:

stock and traditional

Explanation:

the iupac says so

5 0
2 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
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