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Semmy [17]
3 years ago
14

the perimeter of a isosceles is 48 meters. the upper base is 12 meters shorter than the lower base x.each leg is 12 meters short

er than the lower base x. each leg is 12 meters shorter than the lower base. write the equation of the perimeter in terms of x.
Chemistry
1 answer:
mrs_skeptik [129]3 years ago
3 0
Squares have equal sides
12/4=3
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for a theoretical yield of 22 g and actual yield of 13 g, calculate the percent yield for a chemical reaction
sammy [17]

Answer:

59.09%

Explanation:

13/22*100=59.09%

8 0
3 years ago
Why is it important for scientists to review and repeat the work of other scientists
guapka [62]

Answer:

Accuracy

Explanation:

It is important for scientists to review the work of other scientists, so they can be sure there are no mistakes or lack of judgement. They repeat to compare results to make hypotheses.

3 0
3 years ago
2. Consider the reaction 2 Cg H18 (4) +250â (9) ⺠16 co, (g) + 18 HâO(g) la How many moles of H20co) are produced, when |--16:1
Sladkaya [172]

Answer :

(a) The moles of water produced are 145.35 moles.

(b) The mass of oxygen needed are 3080.8 grams.

<u>Solution for part (a) : Given,</u>

Moles of C_8H_{18} = 16.15 moles

First we have to calculate the moles of H_2O

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react to give 18 moles of H_2O

So, 16.15 moles of C_8H_{18} react to give \frac{16.15}{2}\times 18=145.35 moles of H_2O

The moles of water produced are 145.35 moles.

<u>Solution for part (b) : Given,</u>

Mass of C_8H_{18} = 878 g

Molar mass of C_8H_{18} = 114 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_8H_{18}.

\text{ Moles of }C_8H_{18}=\frac{\text{ Mass of }C_8H_{18}}{\text{ Molar mass of }C_8H_{18}}=\frac{878g}{114g/mole}=7.702moles

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

From the balanced reaction we conclude that

As, 2 moles of C_8H_{18} react with 25 moles of O_2

So, 7.702 moles of C_8H_{18} react with \frac{7.702}{2}\times 25=96.275 moles of O_2

Now we have to calculate the mass of O_2.

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(96.275moles)\times (32g/mole)=3080.8g

The mass of oxygen needed are 3080.8 grams.

6 0
3 years ago
Calculate the volume of 38.0 g of carbon dioxide at STP. Enter your answer in the box provided. L
Free_Kalibri [48]

Answer:

19.3 L

Explanation:

V= n × 22.4

where V is volume and n is moles

First, to find the moles of CO2, divide 38.0 by the molecular weight of CO2 which is 44.01

n= m/ MM

n= 38/ 44.01

n= 0.86344012724

V= 0.86344012724 × 22.4

V= 19.3410588502 L

V= 19.3 L

7 0
3 years ago
What do radio waves and microwaves have in common?
Anastasy [175]
1) A
2)A
3)C
4)A
That should be all of them. Hope this helps?
6 0
3 years ago
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