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dimaraw [331]
3 years ago
14

Give the answer to the problem below using the correct number of significant digits. (1.3 x 103) x (5.724 x 104)

Chemistry
2 answers:
Gemiola [76]3 years ago
6 0
(133.9) x (595.296)= 79,710.1344
Answer with significant digits: 80,000
1.3 has two significant digits so 79,700 rounds up to 80000.
Veronika [31]3 years ago
3 0

79710.1344

(1.3 x 103) x (5.724 x 104) = 79710.1344

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NaoH= sodium hydroxide

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A chemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. The concentration
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3.65 g / ml correct to 3 sig. fig.

Explanation:

The computation of the concentration required is shown below:

As we know that

[A] = mass of solute ÷ volume of solution

Before that first find the mass of solute

Given that

Initial weight = 5.55g

And,

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So,

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= 87.15 g ~ 87.2 g

Now the KCi is fully dissolved, so the volume is 23.9 ml

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3 years ago
Charles's law describes the relationship of the volume and temperature of
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The temperature would be reduced by half

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Therefore, if V₁ is reduced to 1/2V₁ = V₂ => V₁/T₁ = 1/2V₁/T₂ => V₁/T₁ = V₁/2T₂

and solving for T₂ =>  1/T₁ = 1/2T₂ => 2T₂ = T₁ => T₂ = 1/2T₁

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Faunal succession and fossil correlation relate to the principle of superposition because there are fossils of certain organisms
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Read 2 more answers
1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
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