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2 years ago

How many meters are there in 23.1 miles?

Chemistry

1 answer:

Hunter-Best [27]2 years ago

4 0

Answer:

3.7 ×10⁴ m

Explanation:

We know that,

1 miles = 1609.34 meter

We need to find how many meters are there in 23.1 miles.

23 miles = 37175.85 m

23 miles = 3.7 ×10⁴ m

So, there are 3.7 ×10⁴ m in 23.1 miles.

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B. Determine the percent composition of each element in NaCl. (1 point) Must show work to

andrey2020 [161]

Answer:

Sodium (Na): (.5 point)

22.99+35.453=58.433. 22.99/58.433= 39%

Chlorine (Cl): (.5 point)

22.99+35.453=58.433. 35.453/58.433= 61%

Explanation:

8 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

2 years ago

A gas expands from a volume of 2.00L at 36.0oC to a volume of 2.50 L, what is the final temperature, if the pressure is constant

lakkis [162]

We shall consider V, the volume and T, the temperature.
According to Boyle's Laws:
$\frac{V1}{T1} = \frac{V2}{T2}$
In our case:
$\frac{2.00}{36.0} = \frac{2.50}{T2} =\ \textgreater \ T2= \frac{2.50*36.00}{2.00} = 45$

6 0

3 years ago

Read 2 more answers

Balance the following

olga55 [171]

Answer:

Your coefficients (the numbers in front of the molecule) will be the following from left to right.

1. 1 - 2 - 1 - 2

2. 2 - 1 - 2 - 2 - 1

3. 2 - 4 - 1

4. 2 - 4 - 3

5. 2 - 2 - 2 - 1

6. 1 - 1 - 1

7. 2 - 1 - 2

8. 3 - 1 - 2 - 3

9. 3 - 1 - 2 - 3

10. 2 - 1 - 1 - 1

Explanation:

To balance this equations first count how many times an element is on each side and then see what needs to be changed in order to balance them.

8 0

7 months ago

. In a separate experiment, the molar mass of nicotine is found to be somewhere between 150 and 180 g/mol. Calculate the molar m

stealth61 [152]

Answer:

Explanation:

The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:

Nicotine has the formula $C_xH_yN_z$ . To determine its composition, a sample is burned in excess oxygen, producing the following results:

1.0 mol of CO₂

0.70 mol of H₂O

0.20 mol of NO₂

Assume that all the atoms in nicotine are present as products

<h2>Solution</h2>

To find the empirical formula you need to find the moles of C, H, and N in each of the compound.

1.0 mol of CO₂ has 1.0 mol of C

0.70 mol of H₂O has 1.4 mol of H

0.20 mol of NO₂ has 0.20 mol of N

Thus, the ratio of moles is:

C: 1.0

H: 1.4

N: 0.20

Divide all by the smallest number: 0.20

C: 1.0 / 0.20 = 5

H: 1.4 / 0.20 = 7

N: 0.20 / 0.20 = 1

Hence, the empirical formula is C₅H₇N

Find the mass of 1 mole of units of the empirical formula:

C: 5mol × 12g/mol = 60g

H: 7mol × 1g/mol = 7 g

N: 1 mol × 14g/mol = 14g

Total mass = 60g + 7g + 14g = 81g

Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.

Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.

5 0

3 years ago