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3 years ago

What hydrocarbons burn completely in an excess of oxygen, the products are

2 answers:

3 0

The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are C) carbon monoxide and carbon dioxide

storchak [24]3 years ago

3 0

Answer:

C just took the test

Explanation:

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While exploring on your grandfathers farm, you find a layer of charcoal that might represent a campfire built by Native American

Ierofanga [76]

Carbon Dating would tell you the age of the charcoal

8 0

3 years ago

How many total atoms are there in one molecule of C145H293O168?

nasty-shy [4]

Answer:

606 atoms

Explanation:

Add the numbers 145 + 293 + 168

4 0

2 years ago

A 2.0 molal sugar solution has approximately the same freezing point as 1.0 molal solution of 1) CaCl2 2) CH3COOH 3) NaCl 4) C2H

sertanlavr [38]

Answer:

3) NaCl.

Explanation:

∵ ΔTf = iKf.m

where, i is the van 't Hoff factor.

Kf is the molal depression freezing constant.

m is the molality of the solute.

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

So, for sugar: i = 1.

∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.

For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.

So, i for NaCl = 2.

∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.

So, the right choice is: 3) NaCl.

8 0

2 years ago

Calculate the percentage yield for the reaction represented by the equation CH4 + 2O2 ? 2H2O + CO2 when 1000 g of CH2 react with

makvit [3.9K]

Answer:

83.64%.

Explanation:

∵ The percent yield = (actual yield/theoretical yield)*100.

actual yield of CO₂ = 2300 g.

We need to find the theoretical yield of CO₂:

For the reaction:

CH₄ + 2O₂ → 2H₂O + CO₂,

1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.

Firstly, we need to calculate the no. of moles of 1000 g of CH₄ using the relation:

no. of moles of CH₄ = mass/molar mass = (1000 g)/(16.0 g/mol) = 62.5 mol.

Using cross-multiplication:

1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.

∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.

We can calculate the theoretical yield of carbon dioxide gas using the relation:

∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.

∵ The percent yield = (actual yield/theoretical yield)*100.

actual yield = 2300 g, theoretical yield = 2750 g.

∴ the percent yield = (2300 g/2750 g)*100 = 83.64%.

5 0

3 years ago

What is the molarity of 555 L of a Ba(OH)2 solution if the pH is 10.20?

Kipish [7]

pH is a symbolic representation of:

$pH=-\log \lbrack H^+\rbrack$

and for water we now that:

$\lbrack H^+\rbrack\lbrack OH^-\rbrack=10^{-14}$

that means the text give us the information to calculate the concentration of protons and therefore the concentration of ions OH-:

$pH=10.2\rightarrow\lbrack H^+\rbrack=10^{-10.2}^{}$

therefore the concentration of OH- can be calculated:

$\lbrack OH^-\rbrack=\frac{10^{-14}}{10^{-10.2}}=10^{-3.8}=1.58\times10^{-4}\text{ M}$

but they are not asking about the concetration of OH- they ask about concentration of Ba(OH)2 and for that we need to know the disolution stechiometry

$Ba(OH)_2\rightarrow Ba^++2\times OH^-$

Whic mean the concentration of Ba(OH)2 is half the concentration of OH-:

$\lbrack Ba(OH)_2\rbrack=\frac{\lbrack OH^-\rbrack}{2}=7.92\times10^{-5}M$

4 0

9 months ago