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STatiana [176]
3 years ago
14

An empty graduated cylinder has a mass of 11.4g.When it is filled with an unknown liquid to the 45.2 mL mark the cylinder and th

e liquid together weigh 95.1 gram. What is the density of the unknown liquid?
Chemistry
1 answer:
Lesechka [4]3 years ago
3 0
The density of the liquid is about 1.85 g/mL.

Density is mass/volume. The volume is given (45.2 mL). The mass must be found by subtracting the tare weight of the graduated cylinder from the total:
95.1 g- 11.4g = 83.7g

Using the mass and volume of the liquid, you can now calculate the density:
d = m/v = 83.7g/45.2 mL = 1.8517699115 g/mL.

Of the original values, the least number of significant figures are 3, so the answer must have a degree of accuracy of 3 significant figures:
1.8517699115 g/mL ≈ 1.85 g/mL.
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1 year ago
Phenol red is an indicator that changes color over a range from ph=6.4 to ph=8.0. what is ka of phenol red?
patriot [66]
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7 0
3 years ago
Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0
3 years ago
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