All categories

4 years ago

Mercury thermometers work because mercury ____ when it is warmed

Chemistry

1 answer:

kap26 [50]4 years ago

5 0

Mercury expands when it is heated. This process is called thermal expansion.

You might be interested in

To completely neutralize a 0.325 g sample of pure aspirin, 15.50 mL of a sodium hydroxide solution is added. If 16.25 mL of the

Kruka [31]

a. 0.325 g / 63.55 g/mol = 5.11 X 10^-3 moles Cu. SHould form 5.11 X 10^-3 mol Cu2+

b. Should form 5.11 X 10^-3 mol Cu(OH)2

c. 1 g Zn / 65.4 g/mol = 0.0153 mol Zn
Excess Zn = 0.0153 - 0.0051 = 0.0102 moles excess zinc

d. 5.11 X 10^-3 mol Mg X 24.3 g/mol = 0.124 grams Mg

3 0

3 years ago

Which of the following molecules is the primary product of photosystem I?a) Carbon dioxideb) Oxygenc) NADPHd) ATP

Lina20 [59]

Answer:

The primary product of the photosystem I is NADPH.

Explanation:

NADPH is the acronym of Nicotinamide adenine Dinucleotide Phosphate. This NADPH plays a vital role in many chemical reactions that take place during photosynthesis process. NADPH is released as a product in Photosystem I.

In the photosystem I process, the molecule present will absorb sunlight energy and transfers it to electrons to produce NADPH. This NADPH molecule is used as fuel for the chemical process that occurs during the 2nd stage of the photosynthesis process. Its molecular formula is $C_{12} H_{30} N_{7} O_{17} P_{13}$ . It acts as a fundamental metabolite.

6 0

3 years ago

Mr.Davies has requested that you wait for 10 more minutes is that informal or formal

LUCKY_DIMON [66]

Answer:

Formal

Explanation:

7 0

4 years ago

Read 2 more answers

Balance this chemical equation.

Natasha_Volkova [10]

Answer:

2 C2H6 + 7 O2 = 4 CO2 + 6 H2O

Explanation:

The Bold Numbers are what you should put. This is balanced

8 0

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)

[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

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] = 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

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

5 0

3 years ago