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Gennadij
3 months ago
9

A flask that can hold 158 g of water at 4oc can hold only 127 g of ethanol at the same temperature. what is the density of ethan

ol?
Chemistry
2 answers:
VMariaS [2.9K]3 months ago
7 0

Answer:

0.8037 g/mL is the density of ethanol.

Explanation:

The mass of the water is m = 158 g.

The volume of water in the flask is V.

The density of water is d = 1.00 g/mL.

\density=\frac{Mass}{Volume}

1.00 g/mL=\frac{m}{V}=\frac{158 g}{V}

V = 158.00 mL

In this case, the ethanol mass held by the flask is M = 127 g.

The volume of the flask measures V = 158.00 mL.

The density of ethanol is D.

D=\frac{M}{V}=\frac{127 g}{158.00 mL}=0.8037 g/mL

0.8037 g/mL is the density of ethanol.

lions [2.9K]3 months ago
3 0

Answer:

  • The density of ethanol measures 0.80 g/ml.

Explanation:

The density of water is known to be 1 g/ml, indicating that 158 g of water occupies a volume of 158 ml.

            d = m/v = 258/258 = 1 g/ml

This implies the volume will remain the same for ethanol.

            hence, the density of ethanol = 127/158 = 0.80 g/ml

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KiRa [2933]
In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>= \frac{weight of CaCO3 (g)}{gram molecular weight}
= \frac{45}{100} = 0.45 mol

Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.

According to the ideal gas equation, we have PV = nRT
V = \frac{nRT}{P}.
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
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5 0
2 months ago
A stock solution of Cu2+(aq) was prepared by placing 0.8875 g of solid Cu(NO3)2∙2.5 H2O in a 100.0-mL volumetric flask and dilut
Anarel [2989]

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

  • The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
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The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

0.8875gCu(NO_{3})_{2}.2.5H_{2}O\times \frac{1molCu(NO_{3})_{2}.2.5H_{2}O}{232.59gCu(NO_{3})_{2}.2.5H_{2}O} \times \frac{1molCu^{2+} }{1molCu(NO_{3})_{2}.2.5H_{2}O} =3.816\times10^{-3} molCu^{2+}

The molarity of Cu²⁺ is:

\frac{3.816\times10^{-3} mol}{100.0 \times10^{-3}L} =3.816\times10^{-2}M

4 0
3 months ago
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