Friction is generated when _____ interfere with each other on sliding surfaces. solids molecules forces fluids

If you added 15,000 calories to 2.0 L of water that was at 25.0 degrees C, what temperature would it be at when you finished?

rewona [7]

Answer:

When we finish, the temperature would be 32.5℃

Explanation:

Density of water = mass/volume

So,

Mass of water = Density × Volume

$\\\\$=1.0 \times 2.0 L$\\\\$=1.0 \frac{g}{m L} \times 2000 m L$\\\\$\quad=2000 g$$

$Q=m \times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

plugging in the values

$15000 \mathrm{Cal}=2000 \mathrm{g} \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times \Delta T$

$\\$\Delta T=\frac{15000 \mathrm{cal}}{2000 \mathrm{g} \times \frac{1.0 \mathrm{cal}}{g^{\circ} \mathrm{C}}}$\\\\$\Delta T=7.5^{\circ} \mathrm{C}$$

Final T = ∆T + Initial T

= 7.5℃ + 25℃ = 32.5℃ (Answer).

5 0

3 years ago

A substance that dissolves another substance is a _____________________

Scorpion4ik [409]

Answer:

b it is easy its b

Explanation:

7 0

3 years ago

Read 2 more answers

24.3 2 An artifact classified as seeds, found in a site at Newlands Cross, Ireland, is found to have a 14C radioactivity of 0.10

Rina8888 [55]

Answer:

Age ≅ 7500 years

Explanation:

All radioactive decay is 1st order kinetics and described by the expression

A = A₀e^-kt => t = ln(A/A₀) / -k

k = 0.693 / t(half life) = (0.693 / 5730)yrs⁻¹ = 1.21 x 10⁻⁴ yrs⁻¹

t = Age = [ln(0.103/0.255) / - 1.21 x 10⁻⁴] yrs = 7500 years

5 0

3 years ago

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

IceJOKER [234]

Answer:

$1.75\cdot 10^{-4} M$

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

$S = k_H p^o$

Where $k_H$ is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

$S_1 = 2.67\cdot 10^{-4} M$

Also, at sea level, we have an atmospheric pressure of:

$p = 1.00 atm$

Given mole fraction:

$\chi_{O_2} = 0.209$

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

8 0

3 years ago

A gas cylinder initially contains 463 L of gas at a pressure of 159 atm. If the final volume of gas is 817 L, what is the final

Agata [3.3K]

Answer:

The final pressure is 90.1 atm.

Explanation:

Assuming constant temperature, we can solve this problem by using Boyle's Law, which states:

P₁V₁=P₂V₂

Where in this case:

P₁ = 159 atm

V₁ = 463 L

P₂ = ?

V₂ = 817 L

We input the given data:

159 atm * 463 L = P₂ * 817 L

And solve for P₂:

P₂ = 90.1 atm

The final pressure is 90.1 atm.

6 0

3 years ago