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Ghella [55]
4 years ago
11

Data: Beaker Test 1:Temp/VolBeaker Test 2:Temp/VolBeaker Test 3:Temp/VolBeaker Test 4:Temp/VolBeaker Test 5:Temp/VolTrial One: A

ir-14.5 °C / 4.3 mL0.9 °C / 4.7 mL21.7°C / 5.0 mL48.5 °C / 5.4 mL81.2 °C / 5.9 mLTrial Two: Nitrogen (N2)-14.5 °C / 4.4 mL0.9 °C / 4.7 mL21.7 °C / 5.1 mL48.5 °C / 5.4 mL81.2 °C / 5.9 mLData Analysis: Create a separate graph of temperature vs. volume for each of the gas samples. You are encouraged to use graphing software or online tools to create the graphs; be sure to take screenshots of the graphs that also include your data.Make sure to include the following on your graphs:•Title•Labels for axes and appropriate scales•Clearly plotted data points
Chemistry
1 answer:
miskamm [114]4 years ago
8 0
A beaker is 1,9005kg because of the size of the weight is 1,000kg=1g
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WILL GIVE BRAINLIEST
Marianna [84]
Answer:

B, C, and D

Explanation:

A is the only one in which two components are being combined. The point is to separate the mixture, so that is the only one that would not apply.
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61.2 grams of hydrogen sulfide react with 64.0 grams of Sulfur dioxide and produce 62.2 grams of solid sulfur (S8). (a) What amo
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Yes Because he really needs it. The amount of sulfur would be
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What is the Molar Mass of methylammonium bromide: Use the Periodic Table and round to the nearest hundredths.
egoroff_w [7]

The molar mass of methylammonium bromide is 111u.

<h3>What is molar mass?</h3>

The molar mass is defined as the mass per unit amount of substance of a given chemical entity.

Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.

Add it all together and put units of grams/mole after the number.

Atomic weight of H is 1u

Atomic weight of N is  14u

Atomic weight of C is  12u

Atomic weight of Br is  79u

Calculating molar mass of  2H_3NCH_3Br =2(1 x3+ 14+12+ 1 x 3 +79) = 111u

Hence, the molar mass of methylammonium bromide is 111u.

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8 0
2 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

5 0
3 years ago
A pure solvent has a boiling point ____ the boiling point of a solution.
Inessa05 [86]
Boiling-point<span> elevation describes the phenomenon that the </span>boiling point<span> of a liquid (a </span>solvent<span>) will be higher when another compound is added, meaning that a </span>solution has<span> a higher </span>boiling point<span> than </span>a pure solvent<span>. This happens whenever a non-volatile solute, such as a salt, is added to </span>a pure solvent<span>, such as water.

So the correct answer is C.</span>
4 0
3 years ago
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