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2 years ago

Which of the following equalities is not correct?

Chemistry

2 answers:

Kryger [21]2 years ago

8 0

1cm^3 = 1L would be the correct answer. One cubic centimeter equals .001 liter, so this equality above is not correct.

Please let me know if you have any questions! :)

tankabanditka [31]2 years ago

8 0

Answer:

10kg=1g

Explanation:

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How many grams are there in 7.5x10^23 molecules of H2SO4

horsena [70]

Moles of H2SO4= 7.5x10^23/ 6.02x10^23 = 1.25 (3sf) moles of H2SO4

Mass of 1 mole of H2SO4= 98.1g

Therefore mass of 7.5x10^23 molecules of H2SO4= 122.63g

6 0

2 years ago

How many liters of 3.5 M solution can be made using 23 moles of LiBr *must show work to get credit*

MrMuchimi

<u>Answer:</u> 6.57 L of solution can be made.

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$ .....(1)

Given values:

Molarity of LiBr = 3.5 M

Moles of LiBr = 23 moles

Putting values in equation 1, we get:

$3.5mol/L=\frac{23mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{23mol}{3.5mol/L}=6.57L$

Hence, 6.57 L of solution can be made.

5 0

2 years ago

30) Which of these statements are true?

Katen [24]

D) 1 and 4
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4 0

2 years ago

Read 2 more answers

In a reaction between 6.0 g of oxygen gas, 4.0 g of hydrogen gas, and 5.0 g of solid sulfur at standard temperature and pressure

nikitadnepr [17]

Answer:

The limiting reagent is the O₂

Explanation:

We can think, this reaction

2O₂(g) + H₂(g) + S(s) → H₂SO₄

Mole of each = Mass / molar mass

6 g / 32 g/m = 0.187 mole O₂

4g / 2 g/m = 2 mole H₂

5g / 32.06 g/m = 0.156 mole S

Ratio between reactants is 2:1:1, 1:2:1, 1:1:2

For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S

0.187 mole of O₂, I need (the half)

0.093 mole of H₂ and 0.093 mole of S

For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S

2 mole of H₂, I need (the double of O₂ and the same for S)

4 mole of O₂ ; 2 mole of S

For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂

0.156 mole I need the same amount for H₂ and the double for O₂

0.156 mole of H₂ and 0.312 mole of O₂

In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.

6 0

2 years ago

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Degger [83]

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4 0

1 year ago