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Solnce55 [7]
2 years ago
8

Calculate the number of moles in 2.67 g of carbon trihydride

Chemistry
1 answer:
Scorpion4ik [409]2 years ago
8 0

Answer:0.178 moles

Explanation: carbon trihydride seems to be an unusual name for the methyl group CH3–

ionic wt 15

moles = 2.67/15 = 0.178

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A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:
Arlecino [84]

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture  = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100

Percent recovery of compound A:

\frac{83 mg}{119 mg}\times 100=69.74\%

Mass of compound B in a mixture  = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

\frac{79 mg}{97 mg}\times 100=81.44\%

3 0
2 years ago
Which methods would be suitable for determining the concentration of an aqueous solution of KMnO4? I. Visible spectrophotometry
Elodia [21]

Answer:

C

Explanation:

Since the solution have an observable color, that means that it absorbs light in the visible region hence it can be determined by colorimetry. Secondly, KMnO4 is a reducing agent which can be titrated against an oxidizing agent and it's concentration accurately determined.

3 0
3 years ago
Write the formulas of the following compounds:
dem82 [27]

Answer:

a) Li2CO3

b) NaCLO4

c) Ba(OH)2

d) (NH4)2CO3

e) H2SO4

f) Ca(CH3COO)2

g) Mg3(PO4)2

f) Na2SO3

Explanation:

a) 2Li + CO3 ↔ Li2CO3

b) NaOH * HCLO4 ↔ NaCLO4 + H2O

c) Ba + 2H2O ↔ Ba(OH)2 +

d)  2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O

c) SO2 + NO2 +H2O ↔ H2SO4 + NOx

f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O

g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O

h) NaOH + H2SO3 ↔ Na2SO3 + H2O

6 0
3 years ago
Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid w
Naddika [18.5K]

Answer:

87.9 % is the percent yield of H₂O

Explanation:

This is the neutralization reaction. A base reacts with an acid to produce water and the correspondly ionic salt.

NaOH  +  HCl  → NaCl +  H₂O

As we have the mass of the two reactants, we must determine the limiting reactant.

Let's convert to moles, the mass of each reactant. (mass / molar mass)

21.1 g / 36.45 g/mol = 0.579 moles of HCl

46.3 g / 40g/mol = 1.15 moles of NaOH

Ratio is 1:1, so it is obviously that the limiting reactant is the HCl. For 1.15 moles of NaOH, i need the same amount of acid, but I only have 0.579 moles

Let's work with the products now. Ratio is 1:1 again, so If I have 0.579 moles of acid, I can produce 0.579 moles of H₂O.

How many grams are 0.579 moles of water? We should find it out as this

mol . molar mass = mass → 0.579 mol . 18 g/mol = 10.4 g

We were told that the production of water was 9.17 g, so let's determine the percent yield as this:

(Yield produced / Theoretical yield) . 100 =

(9.17 g / 10.4g ) . 100 = 87.9 %

6 0
3 years ago
A student dissolves equal amounts of salt in equal amounts of warm water, room-temperature water, and ice water. Which result is
Tanya [424]

doesnt salt desolve ice? so wouldn't the salt dissolve in the ice water?

6 0
2 years ago
Read 2 more answers
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