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2 years ago

What is the molar mass of a compound if 25.0 g of the compound dissolved in 750. Ml gives a molarity of 0.290 m?

Chemistry

1 answer:

wlad13 [49]2 years ago

6 0

Answer:

115g/mol

Explanation:

To get the molar mass, we know that the it is equal to the mass divided by the number of moles. We have the mass but we do not have the number of moles.

We get this by working through the solution information. Firstly, we need to know the number of moles in 750ml for a molarity of 0.29m

Now, since 0.29 moles is present in 1000ml, x moles will be present in 750ml

The value of x is obtained as follows:

x = (750 * 0.29)/1000 = 0.2175 moles

Now since we have the number of moles, we can then obtain the molar mass.

Molar mass = mass/number of moles = 25.0g/0.2175 = 114.94 approximately 105g/mol

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What else is produced during the combustion of butane, C4H10?

denpristay [2]

Another product: CO₂

<h3>Further explanation</h3>

Given

Reaction

2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O

Required

product compound

Solution

In the combustion of hydrocarbons there can be 2 kinds of products

If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product

If Oxygen is low, you'll get Carbon monoxide(CO) and water

Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products

if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left

2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O

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1 year ago

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Which of the following accurately represents the relationship between ceramic and metal?

evablogger [386]

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2 years ago

A 7.591-9 gaseous mixture contains methane (CH4) and butane

mestny [16]

Answer:

65.71%

Explanation:

First, we can write the mass of the mixture, thus:

7.519g = X + Y <em>(1)</em>

<em>Where X is the mass of methane and Y the mass of butane</em>

<em />

Also, the reactions of combustion are:

CH₄ + 2O₂ → CO₂ + 2H₂O

<em>2 moles of oxygen react per mole of methane</em>

C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O

<em>13/2 moles of oxygen react per mole of methane</em>

<em />

That means, in therms of moles of oxygen we can write:

0.9050 moles = 2X/16.04 + 13/2Y/ 58.12

0.9050 = 0.12469X + 0.11184Y <em>(2)</em>

<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>

That is because if X is the mass of methane:

X g Methane * (1mol / 16.04g) = Moles methane

Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen

Replacing (1) in (2):

0.9050 = 0.12469X + 0.11184 (7.519 - X)

0.9050 = 0.12469X + 0.841 - 0.11184X

0.0641 = 0.01285X

X = 4.988g = Mass of methane.

And mass percent of methane is:

4.988g / 7.591g * 100

7 0

2 years ago

Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide

Rama09 [41]

<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

= 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

= 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>

We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

$Percent yield=(\frac{Actual yield}{theoretical yield})100$

$Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100$

= 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

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2 years ago

x g of metal (equivalent wt=12) was completely dissolved in 100 cc of N/2 HCl. The volume was then made upto 500 cc.25 cc of thi

ad-work [718]

Answer:

zxzx

Explanation:

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2 years ago