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jeyben [28]
3 years ago
9

30 POINTS!!!!!! PLEASE HURRY!!!! Four boxes of increasing size and weight must be pushed up a ramp. To push the boxes at the sam

e speed, what must you do? A) Push the boxes with the same force, smoothly up the ramp.. B) The larger and heavier the box, the more force you must use when pushing. C) Push the boxes with the same force, at exactly the same speed, up the ramp. D) Push the largest box with the most force; push the others with about the same force.
Chemistry
2 answers:
Verdich [7]3 years ago
8 0

Answer:

The larger and heavier the box, the more force you must use when pushing.

Explanation:

faust18 [17]3 years ago
7 0

The larger and heavier the box, the more force you must use when pushing.

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What is a precipitate?
Mariulka [41]
Answer: 1) A solid product of a chemical reaction that is in aqueous form.
6 0
3 years ago
What is the name of the compound with the formula PbF2?
Ber [7]

Lead fluoride hope this helps

4 0
3 years ago
Calculate each of the following quantities.<br> (a) mass in kilograms of 3.7 x 1020 molecules of NO2
HACTEHA [7]

Answer:

The answer to your question is:  0.028 kg of NO2

Explanation:

Data

3.7 x 10²⁰ molecules of NO2 in kg

MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg

                   1 mol of NO2 ---------------------  6.023 x 10 ²³ molecules

                   x                     --------------------- 3.7 x 10²⁰ molecules

                   x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

                   x = 0.00061 mol

         

                 1 mol of NO2 ---------------------  46 kg of NO2

                 0.00061 mol     ------------------    x

                 x = 0.00061 x 46/1

                x = 0.028 kg of NO2

7 0
3 years ago
In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i
GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: V_{1} = 21.35 mL,        M_{1} = 0.150 M

V_{2} = 25.0 mL,           M_{2} = ?

Formula used to calculate molarity of H_{2}SO_{4} is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = V_{1} + V_{2}

= 21.35 mL + 25.0 mL

= 46.36 mL  (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol

Thus, we can conclude that there are 0.006 moles of acid in the flask.

3 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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