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3 years ago

How many sig figs are in 0.00205kg

Chemistry

2 answers:

gavmur [86]3 years ago

6 0

Answer:

Three

Explanation:

Significant figures:

The given measurement have three significant figures 2,0,5.

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

fiasKO [112]3 years ago

4 0

It’s 20500000, here are some extras if u need

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3 years ago

Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

Igoryamba

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

The pH of the solution is 3.15.

Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

c 0 0

Equilibrium

c-x x x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$ $K_{a2}=5\times 10^{-12}$

Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

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