Dissolving NaOH (s) in water is exothermic. Two calorimetry experiments are set up. Exp 1: 2 g of NaOH are dissolved in 100 mL of water

Exp 2: 4 g of NaoH are dissolved in 200 mL of water.

Which of the following is true?

A. both temperature changes will be the same
B. the second temeprature change will be approximately twice the first
C. the second temperature change will be approximately four times the first
D. the second temperature change will be approximately one-half of the first
E. the second temperature change will be approximately one-fourth the first

Option-E

Explanation:

In the given question, the experiment was set up with 2 reactions in which reaction two has double the amount of reactants.

To calculate the heat released in experiments, the formula used will be

ΔH = mcΔT, where m= mass of

1. For reaction 1,

ΔH₁ = m₁c₁ΔT₁

2. For reaction 2,

ΔH₂ = m₂c₂ΔT₂

here, the concentration of each solute is the same but the amount is different. Reaction two has double the amount compared to the first reaction.

we compare ΔH₁ = 2 ΔH₂

m₁c₁ΔT₁=  m₂c₂ΔT₂

here since concentration is the same, therefore cancelled out

m₁ΔT₁ =  2 m₂ΔT₂

m₁/2 x m₂ xΔT₁ = ΔT₂

100 / 2 x 200 ΔT₁

= 100/400 ΔT₁

= 1/4  ΔT₁= ΔT₂.

This shows that the second temperature change is 1/4 of the first.

Thus, Option-E is the correct answer.

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

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chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.