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3 years ago

Please help me with this answer correct

Chemistry

2 answers:

IRINA_888 [86]3 years ago

8 0

Trust in God and he will guide you throughout life, trust me without God life sucks and is meaningless! Jesus was there for you when other hurt and abandoned you he was with you comforting you, don’t neglect him trust and love him! Repent of your sins and live by the Bible!

oksano4ka [1.4K]3 years ago

5 0

Answer:

hope this helps :)

Explanation:

for the first one, you can look at the periodic table and look at the atomic number and it will show you how many protons there are giving you the answer because protons and electrons are equal in a pure element

a- carbon

b- neon

c- boron

d- oxygen

e- helium

f- hydrogen

g- lithium

h- beryllium

i- nitrogen

1- sulfur

2- S

3- 16

4- 32.066

5- 16

6- 16

7- 16.066

8- draw circles and put 16 dots like on the other page and in the middle put 16 nuetrons and electrons

9- 6 i think

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3 years ago

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;

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<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

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1 year ago

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Explanation:

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