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3 years ago

Determine how many moles of O2 are required to react completely with 4.6 mol of C4H10

Chemistry

1 answer:

kondaur [170]3 years ago

3 0

Answer:

29.9 moles

2C²H¹⁰ needs 13 moles of O²

4.6 C⁴H¹⁰ needs X moles of O²

X= 13× 4.6 ÷ 2 = 59.8 ÷ 2 = 29.9 moles

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The solubility of a solute and solvent at a given temperature is 30.65 g/100 mL.

Alisiya [41]

Answer:

613.0 g

Explanation:

2 L = 2000 mL

(30.65 g/100 mL)*2000 mL = 613.0 g

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3 years ago

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Why do some elements that are classified as metals not look metallic?

Shkiper50 [21]

Answer:

So, that students like us get confuse!!!

Explanation:

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3 years ago

What is the molarity of an intravenous glucose solution prepared from 108 g of glucose in 2.0 L of solution? 0.018 mol/L 0.30 mo

scZoUnD [109]

Answer:

0.30 mol/L

Explanation:

Mass = 108 g

Molar mass of glucose = 180.156 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{108\ g}{180.156\ g/mol}$

$Moles= 0.5995\ mol$

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3 years ago

What elements does the compound iron oxide (Fe2O3)have

Sergeu [11.5K]

2 Iron atoms and 3 oxygen atoms

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3 years ago

Calculate the volume in liters of a 29.8 g/dL nickel(II) chloride solution that contains 131. G of nickel(II) chloride . Be sure

ohaa [14]

Answer:

0.44 L.

Explanation:

Density of nickel(II) chloride = 29.8 g/dL.

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Next, we shall convert 29.8 g/dL to g/L. This can be obtained as follow:

Recall:

1 g/dL = 10 g/L

Therefore,

29.8 g/dL = 29.8 x 10 = 298 g/L

Therefore, 29.8 g/dL is equivalent to 298 g/L.

Finally, we shall determine the volume of nickel(II) chloride as follow:

Density of nickel(II) chloride = 298 g/L

Mass of nickel(II) chloride = 131 g

Volume of nickel(II) chloride =?

Density = mass /volume

298 = 131/Volume

Cross multiply

298 x Volume = 131

Divide both side by 298

Volume = 131/298

Volume = 0.44 L

Therefore, the volume of of nickel(II) chloride is 0.44 L

3 0

3 years ago