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maxonik [38]
3 years ago
10

Convert the density of water, 1.00 g/mL to lb/ft3. Provide only the numerical value in your answer with the correct number of si

gnificant figures.
Chemistry
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

The density of water 1.00 g/ml=62.43 lbs/ft^3.

Explanation:

Density of water = 1.00 g/mL

1 lb = 453.592 g

1 g=\frac{1}{453.592} lbs

ft^3=28316.8 mL

1 mL=\frac{1}{28316.8} ft^3

Density of the water in lb/ft^3:

\frac{1.00 g}{1 mL}=\frac{1.00}{453.592 } lb\times \frac{28316.8 }{1 ft^3}

=62.43 lbs/ft^3

The density of water 1.00 g/ml=62.43 lbs/ft^3.

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Why do you think the location surrounding the pacific ocean is known as the ring of fire
snow_tiger [21]
It is on a plate boundary so there are a lot of volcanoes in that area. All the volcanoes form a "ring" around the plate boundary.
6 0
3 years ago
Pls help me i dont know this
r-ruslan [8.4K]

Answer:

1. 12.6 moles

2. 8.95 moles

3.  2A + 5B → 3C

4. 48 moles

Explanation:

1. 2Fe + 3Cl₂ → 2FeCl₃

We assume the chlorine in excess. Ratio is 2:2

2 moles of Fe, can produce 2 moles of chloride

12.6 moles of Fe will produce 12.6 moles of chloride.

2. 2Fe + 3Cl₂ → 2FeCl₃

For the same reaction, first of all we need to convert the mass to moles:

500 g . 1mol / 55.85 g = 8.95 mol

As ratio is 2:2, the moles we have are the same, that the produced

4. The reaction for the combustion is:

2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)

We assume the oxygen in excess.

Ratio is 2:6, so 2 mol of ethane produce 6 moles of water

Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles

3 0
3 years ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
3 years ago
Please help, I lost my previous tutor just when she was explaining the answer
TEA [102]

Answer:

HCO₂/H₂O is not the acid-base conjugate pair.

Explanation:

<em>Acid and conjugate base pairs differ by an H+ ion.</em>

Neither HCO₂ nor H₂O has lost or gained protons.

The conjugate acid of H₂O is H₃O⁺

The conjugate base of HCO₃⁻ is CO₃²⁻

[A conjugate acid has one more H⁺ than its base]

5 0
2 years ago
How many grams are in 3.14 x 1015 molecules of CO?
Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15        Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x                        Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0
3 years ago
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