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3 years ago

6

Two balls are at the same time height and released at the same time. One ball is dropped and hits the ground 5 s later. The othe

r initially moves horizontally. When does the second ball hit the ground? How far does it travel horizontally?

1 answer:

marusya05 [52]3 years ago

7 0

Answer:the other ball hits the ground

15 seconds later

Explanation:

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How do you do stoich

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In a chemical reaction, matter can neither be created nor destroyed, so the products that come out of a reaction must equal the reactants that go into a reaction. Stoichiometry is the measure of the elements within a reaction.[1] It involves calculations that take into account the masses of reactants and products in a given chemical reaction. Stoichiometry is one half math, one half chemistry, and revolves around the one simple principle above - the principle that matter is never lost or gained during a reaction. The first step in solving any chemistry problem is to balance the equation.
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4 years ago

The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)

Fed [463]

Explanation:

(a) Formula that shows relation between $K_{c}$ and $K_{p}$ is as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

Here, $\Delta n$ = 1

Putting the given values into the above formula as follows.

$K_c = K_p \times (RT)^{-\Delta n}$

= $1.20 \times (RT)^{-1}$

= $\frac{1.20}{0.0820 \times 1073}$

= 0.01316

(b) As the given reaction equation is as follows.

$CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

As there is only one gas so ,

$p[CO_{2}] = K_{p}$ = 1.20

Therefore, pressure of $CO_{2}$ in the container is 1.20.

(c) Now, expression for $K_{c}$ for the given reaction equation is as follows.

$K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}$

= $\frac{x \times x}{(a - x)}$

= \frac{x^{2}}{(a - x)}[/tex]

where, a = initial conc. of $CaCO_{3}$

= $\frac{22.5}{100} \times 9.56$

= 0.023 M

0.0131 = $\frac{x^{2}}{0.023 - x}$

x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

% of $CaCO_{3}$ remained = $(\frac{0.017}{0.023}) \times 100$

= 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

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4 years ago

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3 years ago

How many moles of sucroseare in 5.25x1029 sucrosemolecules?[?]x10[?]Enter your answer with the correctnumber of significant figu

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To calculate the number of moles we will use Avogadro's number, which relates the number of molecules contained in a mole of any substance, the relationship between moles and molecules is as follows:

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In the statement we are given the molecules with 3 significant figures, therefore the answer must also have 3 significant figures. So the answer will be: In 5.25x10^29molecules are 8.72x10^5 mol of sucrose

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1 year ago