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faust18 [17]
2 years ago
7

Which statement about gases is true?

Chemistry
2 answers:
Whitepunk [10]2 years ago
7 0

Answer:

they are made up of hard spheres that are in random motion

Romashka-Z-Leto [24]2 years ago
5 0

Answer:

D) They are made up of hard spheres that are in random motion.

Explanation:

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Is a theory testable, yes or no?
Paraphin [41]
Yes because a theory is based on results and the results are part of the experiment and it being tested. You have to test the experiment and get results so yes a theory is testable.
5 0
3 years ago
How many moles are in 110 grams of nahco3
yuradex [85]
Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol

1 mole ---------- 84 g
? mole ---------- 110 g

moles NaHCO₃ = 110 . 1 / 84

moles NaHCO₃ = 110 / 84

= 1.309 moles

hope this helps!
4 0
3 years ago
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0
3 years ago
How many formula units are contained in 2.35mol of Iron (III) Oxide
konstantin123 [22]
Remember, 1 mole= 6.022x10^23 atoms, molecules, or formula units.

Answer is 1.42x10^24

7 0
3 years ago
Write the concentration equilibrium constant expression for this reaction.
Akimi4 [234]

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.  

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]


8 0
3 years ago
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