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Fed [463]
3 years ago
15

Is hot chocolate a pure substance or a mixture? Explain.

Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

It is a mixture

Explanation:

Because it mixes chocolate, milk, sugar, and vanilla to make it sweet and not taste like plain cocoa powder.

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The name for Na2 O in chemistry ​
IgorC [24]
Pretty sure it’s sodium oxide.. ?
5 0
3 years ago
Ill give brainliest if you answer right
Paraphin [41]

Answer:

It'll be the same as 58.44 amu. This is because of the law of conservation of mass which states that total mass of reactants is always equal to the total mass of products formed.

pls give brainliest for the answer

3 0
2 years ago
Read 2 more answers
Draw a transition state for the reaction between ethyl iodide and sodium acetate
SCORPION-xisa [38]
Transition state <span> for the reaction between ethyl iodide and sodium acetateis shown below </span>

6 0
3 years ago
To determine the concentration of an EDTA solution, 4.11 g of Zn metal was used. The volume of EDTA solution needed to reach the
USPshnik [31]

Answer:

2.23M

Explanation:

Molarity of a solution is calculated thus

Molarity = number of moles (n) ÷ volume (V)

According to this question, 4.11g of Zn metal was used in order to reach a volume of EDTA solution of 28.26 mL.

28.26mL = 28.26/1000

= 0.02826L

Using mole = mass/molar mass to calculate no. of moles of Zn

Mole = 4.11/65.4

mole = 0.0628mol

Molarity = 0.0628 ÷ 0.02826

Molarity = 2.23M

The concentration of the EDTA solution used is 2.23M

7 0
3 years ago
How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
Dima020 [189]

Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

Best regards.

4 0
3 years ago
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