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kotegsom [21]
3 years ago
14

What is the percent by mass of nahco3 in a solution containing 10 g of nahco3 dissolved in 400 ml of h2o?

Chemistry
1 answer:
Marat540 [252]3 years ago
7 0
Answer is: the percent by mass of NaHCO₃ is 2,43%.
m(NaHCO₃) = 10 g.
V(H₂O) = 400 ml.
d(H₂O) = 1 g/ml.
m(H₂O) = V(H₂O) · d(H₂O).
m(H₂O) = 400 ml · 1 g/ml.
m(H₂O) = 400 g.
m(solution) = m(H₂O) + m(NaHCO₃).
m(solution) = 400 g + 10 g.
m(solution) = 410 g.
ω(NaHCO₃) = 10 g ÷ 410 g · 100%.
ω(NaHCO₃) = 2,43 %
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3 years ago
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Firdavs [7]
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3 0
4 years ago
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Solve and round to the correct number of significant figures: 129 ÷ 29.20
Scilla [17]

4.42

Because when you divide 129/29.20, you get a long string of numbers. 4.417808219178082...

So you round to the significant figure which in this case is 2 decimal places because 29.20 has 2 decimal places.

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5 0
3 years ago
Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

Molarity=\frac{Moles}{Volume(L)}

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

n=0.0366 M\times 5 mol=0.183 mol

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}

106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0
3 years ago
If 60. liters of hydrogen gas at 546 K is cooled to 273 K at constant pressure, the new volume of the gas would be
Korvikt [17]

Answer:30 L

Explanation:

Initial Volume

=

V

1

=

60

l

i

t

e

r

Initial Temperature

=

T

1

=

546

K

Final Temperature

=

T

2

=

273

K

Final Vloume

=

V

2

=

?

?

Sol:-

Since the pressure is constant and the question is asking about temperature and volume, i.e,

V

1

T

1

=

V

2

T

2

⇒

V

2

=

V

1

⋅

T

2

T

1

=

60

⋅

273

546

=

60

2

=

30

l

i

t

e

r

⇒

V

2

=

30

l

i

t

e

r

Hence the new volume of the gas is

30

l

i

t

e

r

6 0
3 years ago
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