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Andrew [12]
3 years ago
12

1. Convert the following temperatures to K. a) 104 C b) -3 C

Chemistry
1 answer:
gogolik [260]3 years ago
3 0

Answer:

a) 377 k

b) 270 k

Explanation:

a) 104 + 273 = 377 k

b) -3 + 273 = 270 k

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For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of
torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For Al_2O_3

Mass of Al_2O_3  = 4.07 g

Molar mass of Al_2O_3  = 101.96 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{4.07\ g}{101.96\ g/mol}

Moles\ of\ Al_2O_3= 0.0399\ mol

According to the reaction:

Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O

1 mole of Al_2O_3  on reaction produces 1 mole of Al_2(SO_4)_3

So,  

0.0399 mole of Al_2O_3  on reaction produces 0.0399 mole of Al_2(SO_4)_3

Moles of Al_2(SO_4)_3  obtained = 0.0399 mole

Molar mass of Al_2(SO_4)_3 = 342.2 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0399= \frac{Mass}{342.2\ g/mol}

Mass= 13.7\ g

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

\%\ yield =\frac{10.4}{13.7}\times 100

<u>% yield =76 %</u>

6 0
2 years ago
Can someone please help me with this I need help asap
DiKsa [7]

Answer:

???

Explanation:

6 0
2 years ago
When would formation of a derivative be a necessary step in this experiment?
const2013 [10]
<span>The formation of a derivative being a necessary step in the experiment lies in the importance of the derived structure. Often the derived product confers to reaction pathways which uses less reactive starting materials and more easily proceeds to completion. This also allows us to take a small amount of sample. The derived product at times is a general compound allowing its easy analysis. Often we encounter a product but we find it difficult to analyse it in ways we want. Here lies the essence of forming a derivative which often are simpler compounds allowing easier analysis yet having similar functional groups and structural properties. Also sometimes we encounter problems when our desired product is unstable and forms stable degraded products. But if we somehow manage to synthesize a derivative it may be relatively stable and form no degradation products. It would be stable at least for a significant period of time making it easier to study its properties. The derived product also at times are synthesized using general reaction pathways facilitating a way of easier synthesis and helping it to correlate with other similar reaction pathways and products.So the above paragraph accounts for the need of derivatives. When we encounter problems similar to those mentioned above it becomes necessary for a researcher to form rather synthesize a derivative.</span>
6 0
3 years ago
´ Identify the independent, dependent, and controlled variables in your experiment which is how does adding salt affect the free
Margaret [11]
<span>An independent variable is a variable that is not affected in the experiment. It is what experimenter controls. Here, the concentration of salt is the independent variable.
On the other side, a dependent variable is called dependent because it depends on the independent variables. It is what is affected and observed during the experiment. Here, the freezing point of water is affected and, thus, it is the dependent variable.
A controlled variable is a variable that must remain unchanged (must be constant) during the experiment so that the effect on the dependent variable depends only on the independent variable. Here, the volume of water must be constant, so it is the controlled variable.

</span>
7 0
3 years ago
Where are reactive metals found?
Charra [1.4K]
On the left of the periodic table
4 0
3 years ago
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