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anastassius [24]
2 years ago
15

Question 18 of 35

Chemistry
1 answer:
sammy [17]2 years ago
4 0

Answer:

A. Dark-colored moths become common after tree trunks become

dark.

Explanation:

I took the test

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1.883 Grams of copper reacted with excess sulfur to form an unknown Copper sulfide. The reaction creates 2.308 grams of an unkno
rosijanka [135]

Answer:

um it is most likely copper

Explanation:

8 0
3 years ago
PLEASE HELP ME
ki77a [65]

Answer:

1.15 atm

Explanation:

According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.

Therefore we have:

Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide

We were given the following:

Total pressure = 2.45 atm

Pressure of oxygen = 0.65 atm

Pressure of carbon monoxide = x

Pressure of carbon dioxide = 0.65 atm

Therefore:

2.45 = x + 0.65 + 0.65

2.45 = x + 1.3

x = 2.45 - 1.3

x = 1.15 atm

6 0
3 years ago
If an unknown element displays many extremely strong metallic properties, where is it most likely to be located on the periodic
dalvyx [7]
It is placed on the far left. Hope I helped:)

3 0
2 years ago
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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
Mendeleev arranged the elements by what physical property? atomic number atomic mass boiling point density
m_a_m_a [10]
Mendeleev arranged the elements according to their atomic masses
3 0
2 years ago
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