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Mice21 [21]
2 years ago
8

Which states of matter change shape when moved? Which states of matter change shape when moved?

Chemistry
1 answer:
lions [1.4K]2 years ago
6 0

Answer:

Gas and Liquids

Explanation:

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Show whether the following solutes are soluble in each solvent, and the process of dissolution of each, and their weak bonds. (N
Alexeev081 [22]

Answer:

I don't really understand this...

5 0
3 years ago
Please help with this , it’s very important
timurjin [86]

Explanation:

a) Cr is <u>acid</u> K is <u>base</u>

b) Ca is <u>base</u> and the other left part is <u>acid</u>

c)Ca is <u>base</u> and F2 is <u>acid</u>

<u>d</u><u>)</u><u> </u>NH4 is <u>acid</u> and SO4 is <u>base</u>

5 0
3 years ago
What are the variables that don't change
Blababa [14]
Constant variable/control variable
7 0
3 years ago
In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant
BaLLatris [955]

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

In another experiment, if 80 XO_3 molecules react with 104 BrZ_3 molecules. How many Br_2 molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of BrZ_3 react with 6 molecule of XO_3

So, 104 molecules of BrZ_3 react with \frac{104}{8}\times 6=78 molecule of XO_3

From this we conclude that, XO_3 is an excess reagent because the given moles are greater than the required moles and BrZ_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of Br_2

From the reaction, we conclude that

As, 8 molecules of BrZ_3 react to give 4 molecules of Br_2

So, 104 molecules of BrZ_3 react to give \frac{104}{8}\times 4=52 molecules of Br_2

Hence, the number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

4 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

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

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
2 years ago
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