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2 years ago

How much heat is needed to raise the temperature of 20.0g of H2O by 10.6oC?

Chemistry

1 answer:

Angelina_Jolie [31]2 years ago

7 0

Answer:

ΔH = 8.87 x 10₅

Explanation:

q = mcΔT = 20g x 1.00 cal/g°C x 10.6°C = 212 cals x 4.184 j/cals = 8.87 x 10⁵ joules

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A reaction has a forward rate constant of 2.3 × 106 s-1 and an equilibrium constant of 4.0 × 108. what is the rate constant for

avanturin [10]

Answer is: 5,75·10⁻¹.
Kf = 2,3·10⁶ 1/s.
K = 4,0·10⁸ 1/s.
Kr = ?
Kf - forward rate constant.
K - equilibrium constant.
Kr - reverse rate constant.
Since both Kf and Kr are constants at a given temperature, their ratio is also a constant that is equal to the equilibrium constant K.
K = Kf/Kr.
Kr = Kf/K = 2,3·10⁶ 1/s ÷ 4,0·10⁸ 1/s = 5,75·10⁻¹.

5 0

3 years ago

List the factor on which Ka of a weak acid depends

lakkis [162]

I don’t know this, so I did research, so here! This is from a website called “Quora”, btw!
If you need more information, just go to this app by copying and pasting your question on the G o o g l e Search bar!

8 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

aksik [14]

Complete Question:

To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?

Answer:

2.23x10⁶ g

Explanation:

The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:

0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³

0.7976 g/m³

The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:

A = πR², where R is the radius = 1.01x10² m (half of the diameter)

A = π*(1.01x10²)²

A = 32047 m²

The volume is then:

V = 32047 * 87.32

V = 2.7983x10⁶ m³

The mass of the F⁻ is the concentration multiplied by the volume:

m = 0.7976 * 2.7983x10⁶

m = 2.23x10⁶ g

8 0

3 years ago

When Rutherford ran his gold foil experiment, he expected to see results like those in the Plum Pudding Atom simulation. Instead

DENIUS [597]

Rutherford's result can not be explained on the basis of the plum pudding model because of the fact that, since some alpha particles were deflected, the atom must contain a small region with a strong electric charge.

The empirical study of the atom led to the emergence of several models of the atom. In the Plum - pudding model, the atom was regarded as a positively charged sphere with embedded negative charges.

This model can not interpret the Rutherford experiment since alpha particles were deflected. The deflection of alpha particles means that, the atom must contain a small region with a strong electric charge.

Learn more: brainly.com/question/730256

4 0

1 year ago