5.) You bought a new car and estimated that your monthly payment would be $312. However, your actual

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

Which of these molecules has an overall dipole moment? O A. F2 B. H2S C. CH4 D. CO2

adelina 88 [10]

Answer:

i think the answer is B cus i think of that

4 0

3 years ago

Study these images

Talja [164]

2. Because a cumulonimbus cloud is a towering vertical cloud

8 0

3 years ago

Read 2 more answers

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

5 0

3 years ago

Identifying Characteristics of the Gas Laws

elixir [45]

Answer: The image from the question has the correct answers.

Explanation:

As summarized in the attached table.

8 0

3 years ago