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KengaRu [80]
2 years ago
5

To investigate the reactivity of metals a student uses four metals. Each time they added 1 g of the metal to 25 cm³ of sulfuric

acid and record the temperature change. Identify the control variables in this example
Chemistry
1 answer:
Mazyrski [523]2 years ago
5 0

Answer:

Volume of the sulfuric acid (25cm³), same mass of each metal (1g)

Explanation:

In an experiment, the CONTROL VARIABLE also known as constant is the variable that is kept unchanged for all groups in an experiment. This is done in order not to influence the outcome of the experiment.

In this case, students are trying to investigate the reactivity of four different metals. They added 1 g of each metal to 25cm³ of sulfuric acid and recorded the temperature change. Based on the explanation of control variable above, the VOLUME OF SULFURIC ACID (25cm³) and the MASS OF EACH METAL (1g) are the CONTROL VARIABLES because they are the same or unchanged in this experiment.

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What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
The teacher said the volume of the liquid was 500 mL when measured a student found it was 499.7 mL what is the students percent
Natalka [10]

Answer:

<h2>0.06 % </h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

error = 500 - 499.7 = 0.3

actual volume = 500 mL

We have

p(\%) =  \frac{0.3}{500}  \times 100 \\  =  \frac{3}{50}  \\

We have the final answer as

<h3>0.06 % </h3>

Hope this helps you

4 0
3 years ago
Is mayonnaise an element compound solution or colloids
Alenkinab [10]
The answer to your question is mayonnaise is colloids.
5 0
3 years ago
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igor_vitrenko [27]

Answer:

I don’t know

Explanation:

5 0
3 years ago
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Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
Read 2 more answers
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