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2 years ago

14

I need to know the independent variable and the dependent variable and control variable

1 answer:

Simora [160]2 years ago

4 0

Independent Variable: a variable that you can change in an experiment

Dependent Variable: something that changes as you change the independent variable

control variable: something that is not changed throughout the experiment

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Does anyone know how to do this???

Bond [772]

Simply look at the periodic table and fill in what you know based on the table

The number of protons = atomic number
The number of electrons, Which is the same as the atomic number for atoms.

The number of valence electrons that is given by the group that the element is in, the top number of each column in the periodic table.

8 0

3 years ago

What two parts of an amino acid are involved in a peptide bond?

Lemur [1.5K]

Answer:

The answer to your question is below.

Explanation:

Amino acids are composed by one amino group, one carboxyl group and one chain.

The parts of the amino acid that are involved in a peptide group are the amino group (- NH₂) and the carboxyl group (-COOH).

6 0

3 years ago

Can you identify 2 characteristics of combustion reactions?

monitta

An element or compound will react with oxygen and will produce carbon dioxide, water, and sometimes carbon (if it is an incomplete combustion).

4 0

2 years ago

This question involves two calculations. The answer to the first part will be

ozzi

Answer :

(a) The mass of $Al_2O_3$ produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of $Al$ = 85.1 g

Molar mass of $Al$ = 27 g/mol

First we have to calculate the moles of $Al$

$\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol$

Now we have to calculate the moles of $Al_2O_3$

The balanced chemical equation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the reaction, we conclude that

As, 4 moles of $Al$ react to give 2 moles of $Al_2O_3$

So, 3.15 moles of $Al$ react to give $\frac{2}{4}\times 3.15=1.58$ mole of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

Molar mass of $Al_2O_3$ = 102 g/mole

$\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g$

Therefore, the mass of $Al_2O_3$ produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%$

Therefore, the percent yield of the reaction is, 72.5 %

8 0

3 years ago

The question is 22 I need help with

romanna [79]

Answer:

i think your best bet is C

Explanation:

6 0

2 years ago