All categories

3 years ago

Ethanol has a density of 0.789 g/cm3. What is the mass of 423 cm3 of ethanol? M = (D)(V)

Chemistry

1 answer:

Ivahew [28]3 years ago

6 0

Answer:

<h2>The answer is 334 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of ethanol = 423 cm³

density = 0.789 g/cm³

So we have

mass = 0.789 × 423 = 333.747

We have the final answer as

Hope this helps you

You might be interested in

Which of these phrases accurately describe disproportionation? Check all that apply.

Olin [163]

Answer:

A single compound is simultaneously oxidized and reduced.

Explanation:

In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.

What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.

Consider the disproportionation of CuCl shown below;

2CuCl -----> CuCl2 + Cu

Here, CuCl2 is the oxidized product while Cu is the reduced product.

7 0

2 years ago

Read 2 more answers

34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i

nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

p(r) = $Ar(r^{2}e^{\frac{-r}{3a_{o}}}$

= $Ar^{3}e^{\frac{-r}{3a_{o}}}$

$\frac{dp(r)}{dr}$ = 0

$Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}}$ + $Ae^{\frac{-r}{3a_{o}}}(3r^{2})$ = 0

$\frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}$

= $3Ar^{2}e^{\frac{-r}{3a_{o}}}$

r = $9a_{o}$

Thus, we can conclude that most likely radius at which the electron would be found is $9a_{o}$ .

4 0

3 years ago

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

8 0

3 years ago

A solution was prepared by mixing 50.0 g

frez [133]

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

4 0

3 years ago

How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4

blsea [12.9K]

Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

C₁V₁=C₂V₂

Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

C₁ = 0.25 M

V₁ = 100 L

C₂ = 7.5 M

V₂ = ?

We input the data:

0.25 M * 100 L = 7.5 M * V₂

V₂ = 3.33 L

Thus the answer is 3.33 liters.

3 0

2 years ago