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spin [16.1K]
3 years ago
13

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

<h3>334 g</h3>

Hope this helps you

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What is the mass of 0.0150 moles of Na₂SO4 (molar mass = 142.04 g/mol)?
olga2289 [7]

Answer:

2.13 g

Explanation:

(142.04)(0.0150) = 2.13 g

4 0
1 year ago
Atoms,which are always neutral in electric charge, contain electrons as well as protons and neutrons. An electron has an electri
creativ13 [48]

Answer:

3

Explanation:

Since the atom has three electrons, it has a -3 electrical charge. Since atoms are neutral, 3 protons are needed to get the charge back to 0.

8 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge=1.6\times 10^{-19}C

1 mole of electrons contain=6.023\times 10^{23} electrons

Thus  1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000Coloumb of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by=\frac{193000}{0.0635}\times 11.5=34952756 Coloumb

Q=I\times t

where Q= quantity of electricity in coloumbs  = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

34952756 C=34.5A\times t

t=1013123sec=281hours

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

8 0
3 years ago
What are the parts of a nucleotide?
kobusy [5.1K]
B-sugar + phosphate + nitrogen base
8 0
2 years ago
Read 2 more answers
Please help...awarding brainliest if correct and lots of points.
abruzzese [7]

Silver chloride produced : = 46.149 g

Limiting reagent : CuCl2

Excess remains := 3.74 g

<h3>Further explanation</h3>

Reaction

silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate

Required

silver chloride produced

limiting reagent

excess remains

Solution

Balanced equation

2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)

mol AgNO3 :

= 58.5 : 169,87 g/mol

= 0.344

mol CuCl2 :

=21.7 : 134,45 g/mol

= 0.161

mol ratio : coefficient of AgNO3 : CuCl2 :

= 0.344/2 : 0.161/1

= 0.172 : 0.161

CuCl2  as a limiting reagent

mol AgCl :

= 2/1 x 0.161

= 0.322

Mass AgCl :

= 0.322 x 143,32 g/mol

= 46.149 g

mol remains(unreacted) for AgNO3 :

= 0.344-(2/1 x 0.161)

= 0.022

mass AgNO3 remains :

= 0.022 x 169,87 g/mol

= 3.74 g

8 0
2 years ago
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