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butalik [34]
3 years ago
9

A baby weighs 7 pounds, 4 ounces at birth and 7 pounds, I ounce at discharge. What percent of weight did the baby lose

Chemistry
1 answer:
Aloiza [94]3 years ago
6 0

Answer:

2.6%

Explanation:

As, 1 ounce (oz) =  0.0625 pounds (lb)

Therefore, weight of baby at discharge = 7 lb,1 oz = 7+0.0625 lb = 7.0625 lb

Since, 1 oz =  0.0625 lb

⇒ 4 oz = 4×0.0625 = 0.25 lb

Therefore, weight of baby at birth = 7 lb,4 oz = 7+0.25 lb = 7.25 lb

The <u>amount of weight lost</u> is equal to the difference of weight of the baby at birth and discharge.

Therefore, <u>weight lost</u> = 7.25 lb - 7.0625 lb = <u>0.1875 lb</u>

Now, the <u>percentage of weight lost</u> by the baby is given by the amount of weight lost divided by the weight of the baby at birth.

Therefore, <u>the percentage of weight los</u>t = weight lost ÷ weight at birth = 0.1875 lb ÷ 7.25 lb × 100 = <u>2.6% </u>

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The  field of Chemistry. 
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What’s a simple machine
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Any of the basic mechanical devices for applying a force, such as an inclined plane, wedge, or lever.
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2 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

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2 years ago
Determine the total time that must elapse until only ¼ of an original sample of the radioisotope Rn-222 remains unchanged.
Arte-miy333 [17]

Answer:

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Explanation:

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Let, the initial amount of the Rn-222 = 1 = A

Final amount = \frac{1}{4} = A'

We will use the following relation for calculating time elapsed in the decay

A' = A(\frac{1}{2} )^\frac{t}{t_1_/_2}  }

Thus,

\frac{1}{4} =1(\frac{1}{2} )^\frac{t}{3.8}

We can write is as,

(\frac{1}{2} )^2=(\frac{1}{2} )^\frac{t}{3.8}

Since the base in both sides are equal, powers can also be equal and thus,

2=\frac{t}{3.8}

So, t = 7.6 days

5 0
2 years ago
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