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Yuri [45]
3 years ago
5

The variables for part 1 of this experiment include the types of materials, used in the cup, the masses of these materials and t

heir temperatures over time. use the drop down menus to complete the sentences and identify the independent and dependent variables.
Chemistry
1 answer:
iVinArrow [24]3 years ago
8 0
So, first off let’s look at what an independent and dependent variable is.

an independent variable doesn’t change when other variables change, therefore making it alone and independent.

a dependent variable changes based on other variables.

so, i don’t have enough information on your experiment to know the answer but think to yourself, what is changing over time based on something else changing (dependent variable) or think does this thing stay the same no matter what (independent variable).

hope this helps!
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1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant
vodomira [7]

The masses of the components are obtained as;

  • Sodium hydrogen carbonate = 3.51 g
  • Sodium carbonate =  8.708 g
<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)

Now, the loss in  mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g -  8.708 g = 1.292 g

Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol

= 3.51 g

Learn more about anhydrous sodium carbonate :brainly.com/question/20479996

#SPJ1

6 0
1 year 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
3 years ago
Which substance is most likely metallic
Montano1993 [528]
I don’t know “which substance” are you talking about??

You can put the picture there so we can see.
8 0
3 years ago
Read 2 more answers
When you stand on a bathroom scale, what force or forces are acting on you?
Readme [11.4K]

Answer:

B

Explanation:

Lets see here, Gravity is pulling you down but your body is repelling to gravity. Meaning that gravity is pulling down your weight and the scale is reading it.

Hope this helped!

3 0
3 years ago
Read 2 more answers
How many moles in 5 grams of gold
Ainat [17]
There is 0.02538502095915 Moles in 5 grams of gold.
8 0
3 years ago
Read 2 more answers
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