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1 year ago

How many grams are contained in a 0.893 mol sample of methane, ch4?

1 answer:

6 0

We are given with a compound, Methane (CH4), with a molar mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in g. We need to solve first the molecular weight of Methane, that is

C=12 g/mol

H=1g/mol

CH4= 12 g/mol +1(4) g/mol = 16 g/mol

With 0.893 mol sample, its corresponding mass is

g CH4= 0.893 mol x 16g/mol =14.288 g

Therefore, the mass of methane is 14.288 g

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How much water must be added to 36.0 g of srcl2 to produce a solution that is 35.0 wt% srcl2? how much water must be added to 36

larisa86 [58]

To solve this problem we will use the following equation:

w = (m of solute) / (m of solution)

w - percentage

It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.

w = mass CaCl2/(mass of water + mass of CaCl2)

mass of water = x

0.35 = 36 / (x + 36)

0.35 × (x + 36) = 36

0.35x + 12.6 = 36

0.35x = 23.4

x = 66.86 g of water is necessary

8 0

1 year ago

The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super

Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0

1 year ago

Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b

frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ ?

(a) $Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}$

(b) $2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}$

(d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Answer: (d) $2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

Explanation: Redox Reaction is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called cell notation, formed by: left side of the salt bridge (||), which is always the anode, i.e., its half-equation is as an oxidation and right side, which is always the cathode, i.e., its half-equation is always a reduction.

For the cell notation: $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$

Aluminum's half-equation is oxidation:

$Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}$

For Lead, half-equation is reduction:

$Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}$

Multiply first half-equation for 2 and second half-equation by 3:

$2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}$

$3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}$

Adding them:

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

The balanced redox reaction with cell notation $Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}$ is

$2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}$

6 0

2 years ago

What is the molarity of 4.35 moles kmno4 dissolved in 750 ml of solution?

Andreyy89

M= #moles / L

4.35/.75 = 5.6

7 0

1 year ago

What is formula for chromium (III) phosphate trihydrate and cobalt (II) phosphate octahydrate

jolli1 [7]

The formula for chromium (III) phosphate trihydrate is CrPO4- 3H20. This compound if in the anhydrous state, exists as a green crystal whereas a hydrated form violet crystal. The formula for cobalt(II) phosphate octahydrate is Co3(PO4)2•8H2O.

7 0

1 year ago

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