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SCORPION-xisa [38]
3 years ago
8

A 1.5 M solution of NaOH was made in a laboratory. If the solution made had a volume of 4.5 L, how many grams of NaOH were added

?
Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

270g

Explanation:

Given parameters:

Concentration of solution = 1.5M

Volume of solution  = 4.5L

Unknown:

Mass of NaOH added  = ?

Solution:

Concentration is the number of moles of solute found in a solution.

 To solve this problem, we need to find the number of moles of NaOH first;

 Number of moles  = concentration x volume

 Number of moles  = 1.5 x 4.5  = 6.75moles

 Now, to find the mass of NaOH;

  Mass of NaOH = number of moles x molar mass

       Molar mass of NaOH  = 23 + 16 + 1  = 40g

 Mass of NaOH  = 6.75 x 40  = 270g

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Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)
marin [14]

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

S.G=\frac{D}{d_w}

d_w = density of water = 1 g/mL

D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = \frac{Mass}{Density}

=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L

1 mL = 0.001 L

Molarity = \frac{n}{V(L)}

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L

5 0
3 years ago
A certain solid has a density of 8.0 g/cm3. if 4.0 g of this solid are poured into 4.00 ml of water, which drawing below most cl
Alexandra [31]
Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
4 0
3 years ago
Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you ne
katrin [286]

Answer:

7.08

Explanation:

To solve this problem we'll use the <em>Henderson-Hasselbach equation</em>:

  • pH = pka + log\frac{[A^-]}{[HA]}

Where \frac{[A^-]}{[HA]} is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

  • pka = -log (1.8x10⁻⁴) = 3.74

We<u> input the data</u>:

  • 4.59 = 3.74 + log\frac{[A^-]}{[HA]}

And<u> solve for </u>\frac{[A^-]}{[HA]}:

  • 0.85 = log\frac{[A^-]}{[HA]}
  • 10^{(0.85)}=\frac{[A^-]}{[HA]}
  • \frac{[A^-]}{[HA]} = 7.08
3 0
3 years ago
2. When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of th
Maksim231197 [3]

Answer:

The answer to your question is C = 1.116 J/g°C

Explanation:

Data

Q = 400 J

mass = 5.6 g

Temperature 1 = T1 = 23°C

Temperature 2 = T2 = 87°C

Specific heat = C = ?

Formula

Q = mC(T2 - T1)

- Solve for C

C = Q / m(T2 - T1)

- Substitution

C = 400 / 5.6 (87 - 23)

- Simplification

C = 400 / 5.6(64)

C = 400 / 358.4

- Result

C = 1.116 J/g°C

7 0
4 years ago
Please help me out!!!
noname [10]
What subject is this ?
6 0
4 years ago
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