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3 years ago

6

Kilogram

2 answers:

yanalaym [24]3 years ago

8 0

Let mass = density × volume
6.28 × 109 ml = 684.52 ml
=0.68452 L = 6.8452 × 10⁻⁴ m³
= 6.8452 × 10⁻³ cubic centimeter
density = 0.1786 kg/cm³
∴ Mass =0.1786 ×( 6.8452 × 10⁻³)
=1.22 × 10⁻³ kg

inysia [295]3 years ago

5 0

Answer:

5 divided by x at x = -1. -Find the derivative of f(x) = negative 7

Explanation:

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3 years ago

Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylam

Vlad1618 [11]

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of $[OH^-]$ be x

Initial concentration of $[CH3NH_2]$ , c = 0.230 M

$CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-$

at eq'm c-x x x

Expression of $K_b$ :

$K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}$

Since ,methyl-amine is a weak base,c>>x so $c-x\approx c$ .

$K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}$

Solving for x, we get:

$x=1.07\times 10^{-2} M$

Given, HCl with 0.230 M , it dissociates fully in water which means $[H^+]$ = 0.230 M

$[OH^-]=[H^+]$ will result in neutral solution, since $[OH^-]$

Remaining $[H^+]$ after neutralizing $[OH^-]$ ions

$[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M$

$pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65$

The pH at the equivalence point for the titration will be 0.65.

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Read 2 more answers