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3 years ago

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6) Sodium bicarbonate can be further broken down into simpler substances by chemical means only. Therefore, sodium bicarbonate i

s a(n)________
a. element
b. compound
c. solution
d. pure substance

2 answers:

Anna35 [415]3 years ago

6 0

Therefore, sodium bicarbonate is a compound

irga5000 [103]3 years ago

3 0

Therefore, sodium bicarbonate is a compound.

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Rain soaks into the soil or runs off into oceans, rivers, and lakes. ___________ which process is this

krok68 [10]

Answer:

The water falls to the earth as precipitation, such as rain, hail, sleet, and snow. When precipitation reaches the earth's surface, some of it will flow along the surface of the land and enter surface water like lakes, streams, and rivers, as runoff. The rest of it soaks or percolates into the soil, called

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3 years ago

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

C)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

D)

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

E)

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

F)

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

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3 years ago

You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25

ICE Princess25 [194]

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of AgCN = n = $\frac{125 g}{134 g/mol}=0.9328 mol$

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

$V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL$

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

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Arte-miy333 [17]

Answer:

I think it will option B it will retain enough heat

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How many moles of Cu2S can be formed from 11 moles of HCl?

natima [27]

Answer:

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Explanation:

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