We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:

KOH + H₂SO₄ → H₂O + KHSO₄

If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:

0.025 L x 0.150 mol/L = .00375 mol KOH

0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄

We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:

0.00375 mol / 0.015 L = 0.25 mol/L

The concentration of H₂SO₄ being neutralized is 0.25 M.

6 0

2 years ago

I need help with question 1 and 3! I don’t get this at all!

velikii [3]

Answer:

1) A transform boundary is a boundary plate in which the motion usually lies horizontal.

3) They can be fount at the end of all costs of the continents

5 0

3 years ago

25 points please help

maxonik [38]

Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.

Solution : Given,

For Accelerator 1 model,

Input energy = 2078.3 J

Wasted energy = 663.1 J

Output energy = 1415.2 J

For Accelerator 2 model,

Input energy = 7690.0 J

Wasted energy = 2337.5 J

Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

Output energy = 1802.3 J

Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy ) × 100

For Accelerator 1 model,

% energy lost as waste = $\frac{663.1}{2078.3}\times100$ = 31.90%

For Accelerator 2 model,

% energy lost as waste = $\frac{2337.5}{7690.0}\times100$ = 30.39%

For Accelerator 3 model,

% energy lost as waste = $\frac{2259.6}{4061.9}\times100$ = 55.62%

So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.

6 0

2 years ago