Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:

When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:

1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid
Answer:
Copper nitrate is a blue coloured crystal.which copper nitrate is heated it loses the water molecule and then when is further heated it decomposes to give redddish brown gas,nitrogen dioxide and oxygen.
Explanation:
Answer:
Rb = +1 , Sr = +2, In= +3, Sn = +4, Sb= +5
Explanation:
Formula:
Zeff = Z - S
Z = atomic number
S = number of core shell or inner shell electrons
For Sn:
Electronic configuration:
Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²
Zeff = Z - S
Zeff = 50 - 46
Zeff = +4
For Rb:
Electronic configuration:
Rb₃₇ = [Kr] 5s¹
Zeff = Z - S
Zeff = 37 - 36
Zeff = +1
For Sb:
Electronic configuration:
Sb₅₁ = [Kr] 4d¹⁰ 5s² 5p³
Zeff = Z - S
Zeff = 51 - 46
Zeff = +5
For In:
Electronic configuration:
In₄₉ = [Kr] 4d¹⁰ 5s² 5p¹
Zeff = Z - S
Zeff = 49 - 46
Zeff = +3
For Sr:
Electronic configuration:
Sr₃₈= [Kr] 5s²
Zeff = Z - S
Zeff = 38 - 36
Zeff = +2
The answer is 1/8.
Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1.

,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample
2.

where:
<span>

- half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
<span>

= 28.8 years
We need:
n = ?
x = ?
</span>
We could first use the second equation, to calculate n:
<span>If:

,
</span>Then:

⇒

⇒

<span>⇒ n ≈ 3
</span>
Now we can use the first equation to calculate the remained amount of the sample.
<span>

</span>⇒

⇒

<span>
</span>