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Makovka662 [10]
3 years ago
15

Question 2 (1 point)

Chemistry
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

35 amu

Explanation:

In an atom, only masses of protons and neutrons are relevant. electrons are so small in size that their masses are negligible.

The mass of 1 proton/neutron is 1amu.

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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}}

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3 years ago
What two quantities must be known to calculate the density of a sample of matter? A) mass and volume B) length and mass C) size
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Answer:

A

Explanation:

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3 years ago
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What is it called when an enzyme changes shape
7nadin3 [17]

Answer: Induced Fit

Explanation: The process wherein the enzymes change their shape is called as induced fit. It is the precision aligning of enzymes essential for catalytic activity which is caused by the binding of the substrate as enzymes possess active sites. Hence, as the substrate approaches the enzymes, the enzyme alters its shape.

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Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

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3 years ago
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