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4 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]4 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]4 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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How many moles of sodium chloride (NaCl) solute are in 315 grams of a 91.0 percent by mass solution?

Anni [7]

Hope this helps you.

8 0

3 years ago

A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca

Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage = 77.77%

Explanation:

Step 1: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

Step 2: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

Step 3: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

Step 6: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

6 0

3 years ago

Graphite and iodine are non metal but they shine why secondry school

Alex_Xolod [135]

It is their properties. it is made up naturally due to the ions

8 0

3 years ago

What is the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml?

Firdavs [7]

Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock $CuSO_4$ solution = 2.5 M

$V_1$ = volume of stock $CuSO_4$ solution = 5 ml

$M_1$ = molarity of diluted $CuSO_4$ solution = ?

$V_1$ = volume of diluted $CuSO_4$ solution = 750 ml

Putting in the values we get:

$2.5\times 5=M_2\times 750$

$M_2=0.017$

Therefore the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

6 0

3 years ago

The reaction of perchloric acid (HClO4) with lithium hydroxide (LiOH) is described by the equation: HClO4 + LiOH → LiClO4 + H2O

dem82 [27]

Answer:

A. 0.35 M

Explanation:

Hello,

In this case, given the volume and concentration of lithium hydroxide and the volume of chloric acid, we can compute the concentration of the neutralized acid by using the following equation:

$n_{acid}=n_{base}\\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\M_{acid}=\frac{V_{base}M_{base}}{V_{acid}} =\frac{46.9mL*0.75M}{100mL}\\ \\M_{acid}=0.35M$

Therefore, answer is A. 0.35 M.

Regards.

3 0

3 years ago