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yanalaym [24]
3 years ago
9

A solution of dispersant is made by taking 15.0 mL of a 50.0 mg/mL solution of Randyne and mixing it with 50.0 mL of water. Calc

ulate the final concentration of the Randyne in this solution, in units of grams per milliliter.
Chemistry
1 answer:
Len [333]3 years ago
8 0

Answer:

The final concentration of the Randyne in grams per milliliter = 0.011 g/mL

Explanation:

As we know

C1V1 = C2V2

C1 and C2 = concentration of solution 1 and 2 respectively

V1 and V2 = Volume of solution 1 and 2 respectively

Substituting the given values, we get -

50 * 15 = X * (15+50)\\X = 11.54 mg/mL

The final concentration of the Randyne in grams per milliliter = 0.011 g/mL

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Answer:

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8 0
3 years ago
The specific heat of iron is 0.46 J/g°C. How many joules will it take to make the temperature of a 150. g bar go up from 25°C to
leva [86]

Answer:

Q = 2.415kj

Explanation:

Q = Mc∆temperature

c = 0.46j/g°c = 0.46kj/kg°c

mass = 150g = 0.15kg

∆temperature = 60 - 25 = 35°c

Q = 0.15 X 0.46 X 35 = 2.415kj

Q = 2.415kj

7 0
3 years ago
19. Tryptophan is an essential _____ that can be found in the genetic code of humans.
Ivan
The answer is B tryptophan is a amino acid
3 0
3 years ago
AlCl3 + NaOH —> Al(OH)3 + NaCl
victus00 [196]

Hey there!

AlCl₃ + NaOH → Al(OH)₃ + NaCl

Balance OH.

1 on the left, 3 on the right. Add a coefficient of 3 in front of NaOH.

AlCl₃ + 3NaOH → Al(OH)₃ + NaCl

Balance Cl.

3 on the left, 1 on the right. Add a coefficient of 3 in front of NaCl.

AlCl₃ + 3NaOH → Al(OH)₃ + 3NaCl

Balance Na.

3 on the left, 3 on the right. Already balanced.

Balance Al.

1 on the left, 1 on the right. Already balanced.

Our final balanced equation: AlCl₃ + 3NaOH → Al(OH)₃ + 3NaCl

Hope this helps!

6 0
4 years ago
If 12.1 kilograms of al2o3(s), 60.4 kilograms of naoh(l), and 60.4 kilograms of hf(g) react completely, how many kilograms of cr
Fynjy0 [20]
Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.
7 0
4 years ago
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