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2 years ago

you push a friend sitting on a swing. she has a mass of 50kg and accelerates at a rate of 4 m/s2. find the force you exerted

2 answers:

6 0

Mass of the friend sitting on the swing = 50 kg
Rate of acceleration = 4 m/s^2
The above information's are already given in the question. We need to utilize these information's in such a manner that the answer to the question can be determined easily.
We already know that
Force = Mass * acceleration.
This formula has to be used to find the force exerted.
Then
Force exerted = Mass of the friend * Acceleration
= (50 * 4) kg m/s^2
= 200 Newton
So the force exerted by me is 200 Newton. I hope the explanation and the procedure is clear to you and you are satisfied with it.

Cloud [144]2 years ago

5 0

F=ma
f-x
m-50kg
a-4m/s^2

x=50kg*4m/s^2
x=200 Newtons

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Answer:

This question is incomplete

Explanation:

This question is incomplete but what you should know is that isopropanol (also referred to rubbing alcohol) has just one functional group. This functional group is called the hydroxyl group (-OH) and it's the reason the compound name ends with "ol". The hydroxyl group can be seen in the structure of the compound (Isopropanol) below

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2 years ago

4. The stockroom contains 1.0 M NaAc (Sodium Acetate), 1.0 M HAc (acetic Acid), distilled water and strong acids and bases. You

Elina [12.6K]

Answer:

Concentrations of HAc and NaAc you need are 0.122M

Explanation:

pKa of acetic acid is 4.75, that means when amount of sodium acetate and acetic acid is the same, pH will be 4.75

Thus, you know [NaAc]i = [HAc]i

Now, using H-H equation, when pH = 3.75:

3.75 = 4.75 + log [NaAc] / [HAc]

0.1 = [NaAc] / [HAc]

10 [NaAc] = [HAc]

Thus, after the reaction [HAc] must be ten times, [NaAc].

Based in the reaction of NaAc with HCl

NaAc + HCl → HAc + NaCl

Moles of HCl added are:

1mL = 0.001L * (10mol /L) = 0.01 moles HCl.

That means moles of both compounds after the reaction are:

[NaAc] = [NaAc]i - 0.01 mol

[HAc] = [HAc]i + 0.01

Replacing these equations with the information you know:

[NaAc] = [NaAc]i - 0.01 mol

10[NaAc] = [NaAc]i + 0.01

Subtracting both equations:

9[NaAc] = 0.02mol

[NaAc] = 0.0022 moles.

Replacing in [NaAc] = [NaAc]i - 0.01 mol

0.0022mol = [NaAc]i - 0.01 mol

0.0122mol = [NaAc]i = [HAc]i

These moles in 100.00mL = 0.1000L:

[NaAc]i = [HAc]i = 0.0122mol / 0.100L =

0.122M

Thus, concentrations of HAc and NaAc you need are 0.122M

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2 years ago

1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore

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Answer :

(1) The theoretical yield of product, $MgO$ is, 1.257 grams.

(2) The percent yield of $MgO$ is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles$

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the reaction, we conclude that

As, 2 mole of $Mg$ react to give 2 mole of $MgO$

So, 0.03142 moles of $Mg$ react to give 0.03142 moles of $MgO$

Now we have to calculate the mass of $MgO$

$\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO$

$\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g$

Theoretical yield of $MgO$ = 1.257 g

Experimental yield of $MgO$ = 0.8922 g

Now we have to calculate the percent yield of $MgO$

$\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100$

$\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%$

Therefore, the percent yield of $MgO$ is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

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2 years ago