Which household items would be considered a base? Select all that apply. *

Hi can someone pls help it would mean the world to me :)

kkurt [141]

Answer:

Explanation:

3 0

3 years ago

How do the internal structures of vascular and plants transport food and water.

mariarad [96]

All vascular plants have vascular tissue which allows the transport of water, nutrients and food between the ground and the body of the plant. So your answer would be that the internal structures of vascular plants transport food and water through their vascular tissue. Hope this helps!<span />

4 0

4 years ago

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the mechanism for a chemical reaction is shown above. which of the following statements about the overall reaction and rate laws

Naddika [18.5K]

The rate equation of reaction is 2H2O2 → 2 H2O + O2 and the rate law for elementary step 1 is rate=k[H2O2][I−] .

The question is incomplete, the complete question is;

Step 1: H2O2 + I− → IO− + H2O

Step 2: H2O2 + IO− →H2O + O2 + I−

The mechanism for a chemical reaction is shown above. Which of the following statements about the overall reaction and rate laws of the elementary reactions is correct?

A) The chemical equation for the overall reaction is 2 H2O2+I−→ 2 H2O+O2+I− , and the rate law for elementary step 2 is rate=k[H2O2][IO−] .

B) The chemical equation for the overall reaction is H2O2+IO−→ H2O+O2+I− , and the rate law for elementary step 2 is rate=k[H2O2]2[IO−] .

C) The chemical equation for the overall reaction is 2 H2O2→2 H2O+O2 , and the rate law for elementary step 1 is rate=k[H2O2][I−] .

D) The chemical equation for the overall reaction is 2 H2O2→2 H2O+O2 , and the rate law for elementary step 1 is rate=k[H2O2]2 .

It is possible that a chemical reaction may not take place in a single reactive encounter. In that case, we can deduce the sequence of steps by which the reaction occurs. Each step is called an elementary reaction.

The overall rate law is the sum of all the elementary reactions after the intermediates have been eliminated. In this case, the specie IO− is an intermediate. Also, I− appears on both sides of the reaction equation and it has to cancel out.

The rate equation of reaction is 2H2O2 → 2H2O + O2 and the rate law for elementary step 1 is rate=k[H2O2][I−] .

Learn more: brainly.com/question/13309369

8 0

3 years ago

Calculate the freezing point of a solution containing 15 grams of kcl and 1650.0 grams of water. the molal-freezing-point-depres

Soloha48 [4]

Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.

4 0

3 years ago

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If 14.5kj of heat were added to 485g of liquid water, how much would its temp increase

OlgaM077 [116]

<span>7.15 degrees C The specific heat capacity of water is 4.1813 J/(g*K). So we have 3 values with the units kJ, g and J/(g*K). We can trivially convert from kJ to J by multiplying by 1000. And we want to get a result with the unit K (degrees Kelvin). So let's do it. First, let's cancel out the g unit by multiplying. 4.1813 J/(g*K) * 485 g = 2027.9305 J/K Now we can cancel out the J unit by dividing. But if we divide by the energy, we'll be left with the reciprocal of K, not K. So instead divide by the J/K unit. So 14500 J / 2027.9305 J/K = 7.150146418 K Rounding to 3 significant figures gives us 7.15 K. And since degrees C and degrees K are the same size, the temperature will increase by 7.15 degrees C</span>

6 0

3 years ago