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Angelina_Jolie [31]
3 years ago
10

Can anyone help me in my chemistry homework?​

Chemistry
2 answers:
sdas [7]3 years ago
7 0

Answer:

what is in your chemistry home work I can try for...

natta225 [31]3 years ago
6 0

Answer:

1: A substance made by mixing other substances together.

2: Solutions, Suspensions, Colloids and Emulsion.

3: A particular kind of matter with uniform properties.

4: A homogenous mixture is that mixture in which the components mix with each other and its composition is uniform throughout the solution. A heterogenous mixture is that mixture in which the composition is not uniform throughout and different components are observed.

5: A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm. Common examples of solutions are the sugar in water and salt in water solutions, soda water, etc. In a solution, all the components appear as a single phase.

6: A solution forms when one substance dissolves in another. The substance that dissolves is called the solute. The substance that dissolves it is called the solvent. For example, polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

7: Common examples of suspension include the mixture of chalk and water, muddy water, the mixture of flour and water, a mixture of dust particles and air, fog, milk of magnesia, etc. Q2. Give the definition of suspension. Ans: A suspension is a heterogeneous mixture of two or more substances.

8: The Tyndall effect is light scattering by particles in a colloid or in a very fine suspension.

Explanation:

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Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has pre
brilliants [131]

Answer:

\large \boxed{\text{122 000 J}}

Explanation:

1. Calculate the energy needed

\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}}  = \text{ 29 120 cal}

2. Convert calories to joules

\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}

8 0
3 years ago
Which of the following correctly describes how to
AlladinOne [14]

Answer:

b

Explanation:

4 0
3 years ago
How many moles of oxygen are needed to completely react with 9.5 g of sodium
maria [59]
<span>4 Na + O</span>₂<span> = 2 Na</span>₂<span>O
</span>
4* 23 g Na --------> 16 g O₂
9.5 g Na ------------> ?

Mass of O₂ = 9.5 * 16 / 4 * 23

Mass = 152 / 92

Mass = 1.6521 g of O₂

Molar mass O₂ = 16.0 g/mol

1 mole O₂ ------------ 16.0 g 
? mole O₂ ------------ 1.6521 g

mole O₂ = 1.6521 * 1 / 16.0

≈ 0.10325 moles of O₂

hope this helps!

8 0
3 years ago
Ethanol is added to gasoline because the oxygen it contains improves gasoline's burning efficiency. Its combustion reaction is g
Angelina_Jolie [31]

Answer:

A) Exothermic

B) (1) -410.5 kJ  (2) -98.1 kcal

C) -976.44 kJ

Explanation:

A) By the value of ΔH (-1236 kJ), we can see that it's negative, it means that the heat is being lost by the reaction, and, because of that, the reaction is exothermic.

B) The molar mass of ethanol is 46.07 g/mol, so, at 15.3 grams, the number of moles is:

n = mass/molar mass

n = 15.3/46.07

n = 0.3321 mol

By the equation

1 mol of ethanol           ----------------------- -1236 kJ

0.3321 mol of ethanol ----------------------- x

By a simple direct three rule:

x = -410.5 kJ

1 kJ --------------- 0.2390 kcal

-410 kJ ---------- y

y = -98.10 kcal

C) The molar mass of water is 18 g/mol, so the number of moles at 42.7 g is:

n = 42.7/18

n = 2.37 moles

By the equation

3 moles of water     ----------------------- -1236 kJ

2.37 moles of water --------------------- x

By a simple direct three rule:

3x = -2929.32

x = -976.44 kJ

5 0
3 years ago
Determine the mass of CO2 formed if 15 grams of C4H10
Svetlanka [38]

Answer:

m_{CO_2}=45gCO_2

Explanation:

Hello there!

In this case, since the combustion of butane is:

C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O

Thus, since the molar mass of butane is 58.12 g/mol and that of carbon dioxide is 44.01 g/mol, we obtain the following mass of CO2 product:

m_{CO_2}=15gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}} *\frac{4molCO_2}{1molC_4H_{10}} *\frac{44.01gCO_2}{1molCO_2} \\\\m_{CO_2}=45gCO_2

Best regards!

4 0
3 years ago
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