Answer:
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<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
From the question;
1 x 10^5 people can fill 1 stadium
We are given, 1 x 10^18 atoms of iron
We are required to determine the number of stadiums that 1 x 10^18 atoms of iron would occupy.
We are going to assume that a stadium would occupy a number of atoms equivalent to the number of people.
Therefore;
One stadium = 1 x 10^5 atoms
Then, to find the number of stadiums that will be occupied by 1 x 10^18 atoms;
No. of stadiums = Total number of atoms ÷ Atoms in a single stadium
= 1 x 10^18 atoms ÷ 1 x 10^5 atoms
= 1 x 10^13 stadiums
Therefore, 1 x 10^18 atoms of iron would occupy 1 x 10^13 stadiums
Answer:
(a) The system does work on the surroundings.
(b) The surroundings do work on the system.
(c) The system does work on the surroundings.
(d) No work is done.
Explanation:
The work (W) done in a chemical reaction can be calculated using the following expression:
W = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants
R and T are always positive.
- If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
- If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
- If Δn(g) = 0, W = 0, which means that no work is done.
<em>(a) Hg(l) ⇒ Hg(g)</em>
Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.
<em>(b) 3 O₂(g) ⇒ 2 O₃(g)
</em>
Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.
<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)
</em>
Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.
<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>
Δn(g) = 2 - 2 = 0. W = 0. No work is done.
Answer:
a. sulfur difluoride SF₂
b. sulfur hexafluoride SF₆
c. sodium dihydrogen phosphate NaH₂PO₄
d. lithium nitride Li₃N
e. chromium(III) carbonate Cr₂(CO₃)₃
f. tin(II) fluoride SnF₂
g. ammonium acetate NH₄(CH₃COO)
h. ammonium hydrogen sulfate NH₄(HSO₄)
i. cobalt(III) nitrate Co(NO₃)₃
j. mercury(I) chloride Hg₂Cl₂
k. potassium chlorate KClO₃
l. sodium hydride NaH
Explanation:
The names give us information about the composition. First, we mention the cation and then the anion. In the formula, we follow the same order. Each part has a charge but the resulting compound is electrically neutral.
