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kramer
2 years ago
10

A solution containing 42% naoh by mass has a density of 1.30 g/ml. what mass, in kilograms, of naoh is in 6.00 l of this solutio

n?
Chemistry
2 answers:
cupoosta [38]2 years ago
4 0

Answer:

3.276 kg

Explanation:

By definition:

Density of solution =  mass of solution / volume of solution

mass of solution = density of solution * volume of solution

Density of solution: 1.30 g/ml = 1.30 kg/l

mass of the solution = 1.3 * 6.00 = 7.8 kg

By definition:

percent by mass = (mass of solute / mass of solution) *100

mass of solute = (percent by mass/100) * mass of solution

mass of solute = 42/100 * 7.8 = 3.276 kg

Aliun [14]2 years ago
3 0
NaOH
Na=23
O=16
H=1
SO,NaOH=23+16+1
NaOH=40g/mol
42% of 40=42/100x40
oxygen is the compound
density=mass/volume
1.30=mass/6
mass=7.8g
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Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

Nickel is oxidized from oxidation number 0 (Ni) to oxidation number +2, hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

2) sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂.

Iron is oxidized from oxidation number 0 (Fe) to oxidation number +2, hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

3) hydrobromic acid with magnesium :

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

Magnesium is oxidized from oxidation number 0 (Mg) to oxidation number +2, hydrogen is reduced from oxidation number +1 to oxidation number 0 (H₂).

4) acetic acid with zinc:

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Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

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7 0
2 years ago
Students working in lab accidentally spilled 17 l of 3.0 m h2so4 solution. they find a large container of acid neutralizer that
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Answer is: 8568.71 of baking soda.

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c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.

n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).

n(H₂SO₄) = 17 L · 3 mol/L.

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7. Two teams are competing in a tug-of-war contest. Team A is pulling at 4000N and Team B is pulling at 4900N is the opposite di
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Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
2 years ago
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<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

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We are required to calculate the percent yield of sodium peroxide;

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<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

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<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

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Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

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<h3>Step 4: Percent yield of Na₂O₂</h3>
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Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

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Therefore, the percentage yield of sodium peroxide is 87.4%

8 0
2 years ago
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