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2 years ago

Ok help it’s for homework

Chemistry

2 answers:

never [62]2 years ago

4 0

Ability to moderate temperatures is not a property of water

cricket20 [7]2 years ago

4 0

Answer:

C. ability to contract upon freezing

Explanation:

Water can regulate temperatures, it can adhere to other substances via adhesion, and it is called the universal solvent because it can dissolve many types of substances.

When water freezes, it actually expands due to the lattice structure it assumes. In ice, there are more secure bonds that maintain the structure.

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The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of

attashe74 [19]

Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

6 0

3 years ago

QUE ES UN ENLACE GLUCOSÌDICO?

Blizzard [7]

Answer:

Explanation:

no se hombre necisito pontos

8 0

3 years ago

Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g

Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

$\frac{Mass}{Volume}$

$Mass = \frac{percent \;of \;ethanol \;X \;density \;of \;ethanol \;+ \\ \;percent \;of \;water X \;density \;of \;water}{100}$

$0.926 = \frac{x X 0.7890 g/mL \;+ (100-x) X 0.9982 g/mL}{100}$

$x= 34.51$ %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

Learn more about the density here:

brainly.com/question/952755

#SPJ1

8 0

2 years ago

The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is -5,074 kJ/mol. This value is the Delta. Hrxn f

Inessa [10]

Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is $\Delta H_{\rm rxn}$ for $\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O$ .

<h3>What is the enthalpy of reaction?</h3>

The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.

The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.

The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.

The combustion of octane can be given as:

$\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O$

Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is $\rm C_8H_{18}\;+\;25\;O_2\;\rightarrow 8\;CO_2\;+\;9\;H_2O$ . Thus, option B is correct.

Learn more about enthalpy of reaction, here:

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6 0

2 years ago

Draw one product that you would expect from the reaction of 1 mol of 1,3-butadiene and 1 mol of h2o, cat. h2so4?

Daniel [21]

1,3-butadiene is the simplest conjugated diene and undergoes 1,4 addition reaction in acidic environment.
Chemical reaction: CH₂=CH-CH=CH₂ + H₂O → CH₃-CH=CH-CH₂-OH.
CH₂=CH-CH=CH₂ - 1,3-butadiene.
CH₃-CH=CH-CH₂-OH - 2-buten-1-ol.
Diene<span> or </span>diolefin<span> is a </span>hydrocarbon<span> that has two </span>carbon double bonds<span>.</span>

3 0

3 years ago