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3 years ago

A rock falls off a cliff and hits the ground after three seconds. What is its velocity right before it hits the ground?

Chemistry

2 answers:

mezya [45]3 years ago

5 0

The correct answer is 29.4 m/s.

The right formula to use for the scenario given in the question is:

Vf = Vi + at

Where: Vf = Final velocity = ?

Vi = Initial velocity = 0

a = Acceleration [due to gravity] = 9.8 m/s

T = Time = 3 s

Note that an object that is falling down from a height has an initial velocity of zero.

Vf = 0 + [9.8 * 3] = 29.4

Therefore, final velocity = 29.4 m/s.

Amiraneli [1.4K]3 years ago

4 0

To solve this kinematics formula use the following equation:

Vf = Vi + at
Vf = 0 + (9.81 m/s^2)(3 seconds)
Vf = 29.43 m/s and or about 29.4 m/s of reported to 3 significant figures.

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2 years ago

0.200 grams of HCl is dissolved in 0.801 grams water. The density of the solution formed is 1.10 g/mL. What is the molarity of t

alina1380 [7]

Answer:

$M=6.03M$

Explanation:

Hello,

In this case, since the molarity is computed by:

$M=\frac{n_{solute}}{V_{solution}}$

Whereas the solute is the hydrochloric acid, we compute the corresponding moles with its molar mass (36.45 g/mol):

$n_{solute}=0.200gHCl*\frac{1molHCl}{36.45gHCl} =0.00549molHCl$

Next, since the solution contains both HCl and water, we compute the volume in liters by using its density:

$V_{solution}=(0.200+0.801)g*\frac{1mL}{1.10g} *\frac{1L}{1000mL} =9.1x10^{-4}L$

Therefore, the molarity turns out:

$M=\frac{0.00549mol}{9.1x10^{-4}L}\\ \\M=6.03M$

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2 years ago

Determine the value for the following reaction. N2(g) + 3H2(g) → 2NH3(g)+ 22,000 cal ΔH =

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2 years ago

Read 2 more answers

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc

pantera1 [17]

Answer:

$5.35 *10^{-4}$ M

Explanation:

Equation for the reaction is as follows:

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

By Applying the ICE Table; we have

$2CO_{(g)}$ + $O_{2(g)}$ ⇄ $2CO_{2(g)}$

Initial x 0.0025 M 0.0010 M

Change 0 0 0

Equilibrium x 0.0025 M 0.0010 M

$K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}$

Given that $K_c = 1.4*10^2$ ; Then:

$1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}$

$1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}$

$3.5 =( \frac{(0.001)}{(x)})^2$

$\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}$

$1.87=\frac{(0.001)}{(x)}$

$(x)= \frac{(0.001)}{1.87 }$

$x = 5.35 *10^{-4}$ M

∴ The equilibrium concentration of CO = $x = 5.35 *10^{-4}$ M

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