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Leokris [45]
3 years ago
15

Pam's family is clearing some land, and there is a large boulder sitting on top of the ground. Even with the entire family pushi

ng on it, they cannot mov
it. Which type of friction exists between the boulder and the ground

Chemistry
1 answer:
SashulF [63]3 years ago
6 0

Answer:

d

Explanation:

I believe the answer to this question is d , I believe this as the question doesn't explain properly the surface under the boulder. it is questionable as the question is not clear though I believe this as there must be friction under the boulder stopping it from moving.

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Round the answer for the following question to the correct number of significant figures: 38,736 km ÷ 4784 km
uysha [10]

Answer:

Explanation:

38736 ÷ 4784 = 8.097

Sig Figs

4

8.097

Decimals

3

8.097

Scientific Notation

8.097 × 10^0

E-Notation

8.097e+0

Words

eight point zero nine seven

4 0
3 years ago
What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm
lesya [120]

<u>Answer:</u> The mass of sucrose required is 69.08 g

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 8.80 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL    (Density of water = 1 g/mL)

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g

Hence, the mass of sucrose required is 69.08 g

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I have no idea I’m sorry I’m doing a test and need to put a question
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