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4 years ago

Q1: What problem did the development of the metric system solve? Write your answer below!

Chemistry

1 answer:

evablogger [386]4 years ago

8 0

Answer:

It solved the issue of depending on too many base units

Explanation:

It was first adopted by the French government. Metric system was developed by Gabriel Mouton of Lyons, he is called the founding father of the metric system. He proposed the decimal system of measurement that was refined by the scientists.

The metric system is simple for making measurements, you just need to know few metric units. There are seven base units in this system of measurement while there are twenty base units in inch-pound system of measurement. Metric system is easier to remember. The metric system follows the decimal number system and it decreases and increases in size by 10.

It has been adopted by all the countries of the world except Myanmar and US.

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3 years ago

If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

ratelena [41]

Answer: The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

Explanation:

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

$SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial: 0.500 0.500 0.005 0.005

At eqllm: 0.500-x 0.500-x 0.005+x 0.005+x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

We are given:

$K_c=2.50$

Putting values in above equation, we get:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of nitrogen dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of sulfur trioxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Equilibrium concentration of nitrogen monoxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

4 0

3 years ago

Q1: What problem did the development of the metric system solve? Write your answer below!​

Q1: What problem did the development of the metric system solve? Write your answer below!