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Finger [1]
3 years ago
8

In a chemical formula or a chemical equation, what is the small number after the element symbol? ​

Chemistry
1 answer:
trapecia [35]3 years ago
6 0

Answer:

It's called a subscript.

Explanation:

Subscripts tell you how many atoms of an element are in a compound.

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Which has a higher frequency electron or a neutron why ?
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The electron has a higher frequency compared to the neutron. It can be explained by the way an electron orbits the nucleus of an atom. 

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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.

8 0
3 years ago
7. In the experiment discussed in the passage, what is the variable?
Arada [10]

Answer:

A variable is anything that can change or be changed. In other words, it is any factor that can be manipulated, controlled for, or measured in an experiment.

7 0
3 years ago
A distillation column is separating 150.0 kmol/h of a saturated liquid mixture that is 30.0 mol% methanol and 70.0 mol% water. T
hoa [83]

Answer:

Explanation:

From the information given:

Feed F = 150.0 kmol/hr

The saturated liquid mixture of the distillation column is X_F = 30%

Reflux ration  = 2.0%

methanol distillate mole fraction X_D = 0.990

recovery of methanol in the distillate = 97.0%

The distillate flow rate D can be determined by using the formula;

D = 0.97 (F* X_F)

D = 0.97 × 150 × 0.3

D = 43.65 kmol/h

The bottom flow rate Balance B  on the column is:

F = D + B

150 = 43.65 + B

B = ( 150 - 43.65 )kmol/h

B = 106.35 kmol/h

The methanol mole fraction in the bottom x_M can be computed by using the formula:

F*X_F = DX_D + BX_B

150(0.3) = 43.65(0.999) + 106.3(X_B)

45 = 43.60635 + 106.3(X_B)

45 -  43.60635 = 106.3(X_B)

1.39365 = 106.3(X_B)

X_B = 1.39365 /  106.3

X_B =  0.013

the fractional recovery of water in the bottoms f is calculated as:

f = \dfrac{B(1-X_B)}{F_X_F}

f = \dfrac{106.35(1-0.013)}{150\times 0.7}

f = \dfrac{106.35(0.987)}{150\times 0.7}

f = 0.99969

3 0
3 years ago
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