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Yakvenalex [24]
3 years ago
12

Agustina was standing on the edge of a cliff of unknown height holding a bowling ball. She wanted to find the height of the clif

f. She worked with Maite to time the how long the bowling took to fall to the bottom of the cliff. Maite timed the drop at 9.4 seconds on each of their 10 trials. The bowling ball started at rest on each of the trials. What distance did the bowling ball fall in each trial?
Chemistry
1 answer:
nlexa [21]3 years ago
5 0

Answer:

433 m

Explanation:

Since the fall represents motion under gravity, we use the equation

s = ut - 1/2gt² where s = height of cliff or distance bowling ball falls through, u = initial velocity of bowling ball = 0 m/s(since it starts from rest), t = time = 9.4 s and g = acceleration due to gravity = -9.8 m/s².

So, substituting the values of the variables into the equation, we have

s = 0 m/s × 9.4 s - 1/2 × 9.8 m/s² × (9.4 s)²

s = 0 m - 1/2 × -9.8 m/s² × 88.36 s²

s = 1/2(865.928 m)

s = 432.964

s ≅ 433 m

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AlekseyPX

Answer:

A) chlorine

Explanation:

To solve this question we can use:

PV = nRT

In order to solve the moles of the gas. With the moles and the mass we can find the molar mass of the gas to have an idea of its identy:

PV = nRT

PV / RT = n

<em>Where P is pressure: 603mmHg * (1atm / 760mmHg) = 0.7934atm</em>

<em>V = 100mL = 0.100L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 14°C + 273.15 = 287.15K</em>

0.7934atm*0.100L / 0.082atmL/molK*287.15K = n

3.37x10⁻³ moles of the gas

In 0.239g. The molar mass is:

0.239g / 3.37x10⁻³ moles = 70.9g/mol

The gas with this molar mass is Chlorine, Cl₂:

<h3>A) chlorine </h3><h3 />

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3 years ago
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Yakvenalex [24]
We can use distillation to separate liquid to liquid
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Please do 1-5! giving 5 stars and thanks if right!
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A steel container with a movable piston contains 2.00 g of helium which was held at a constant temperature of 25 °C. Additional
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Answer: D) 1.00 g

Explanation:

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

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\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = 2.00 L

V_2 = final volume of gas = 3.00 L

n_1 = initial moles of gas  =\frac{\text {Given mass of helium}}{\text {molar mass of helium}}=\frac{2.00g}{4g/mol}=0.500mol

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{2.00L}{3.00L}=\frac{0.500mol}{n_2}

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