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makvit [3.9K]
2 years ago
9

Need help asap!!!! will mark brainliest!!!

Chemistry
1 answer:
SOVA2 [1]2 years ago
7 0

Answer:

picture not bright

Explanation:

please a brainliest for tge feedback

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The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
Mademuasel [1]

Answer:

\large \boxed{\textbf{609 kJ}}  

Explanation:

The formula for the heat absorbed is

q = mCΔT

Data:

m = 2.07 kg

T₁ = 23 °C

T₂ = 191 °C

C = 1.75 J·°C⁻¹g⁻¹

Calculations:

1. Convert kilograms to grams

2.07 kg = 2070 g

2. Calculate ΔT

ΔT = T₂ - T₁ = 191 - 23  = 168 °C

3. Calculate q

q = \text{2070 g} \times 1.75 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 168 \,^{\circ}\text{C} = \text{609 000 J} = \textbf{609 kJ}\\\text{The heat energy required is }\large \boxed{\textbf{609 kJ}}

4 0
3 years ago
Ethanoic acid, CH3COOH, ionizes to form an ethanoate ion (CH3COO-) in aqueous solution. what else does it form? is it a monoprot
barxatty [35]

Ethanoic acid ionizes in aqueous solutions to form two ions which are CH_3COO^- and H^+

<h3>Ionization of ethanoic acid</h3>

Ethanoic acid goes by the chemical formula CH_3COOH.

In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

CH_3COOH ---- > CH_3COO^- + H^+

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single H^+

More on ethanoic acid can be found here: brainly.com/question/9991017

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8 0
2 years ago
A 12.0% sucrose solution by mass has a density of 1.05 gem, what mass of sucrose is present in a 32.0-mL sample of this solution
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Answer:

Option C. 4.03 g

Explanation:

Firstly we analyse data.

12 % by mass, is a sort of concentration. It indicates that in 100 g of SOLUTION, we have 12 g of SOLUTE.

Density is the data that indicates grams of solution in volume of solution.

We need to determine, the volume of solution for the concentration

Density = mass / volume

1.05 g/mL = 100 g / volume

Volume =  100 g / 1.05 g/mL → 95.24 mL

Therefore our 12 g of solute are contained in 95.24 mL

Let's finish this by a rule of three.

95.24 mL contain 12 g of sucrose

Our sample of 32 mL may contain ( 32 . 12) / 95.24 = 4.03 g

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