What happens when Identical objects are dropped at different gravitational forces

What is the difference in mean solar time between 30 N 75 W and 30 N 90 W

ratelena [41]

Answer:

Every 15 degrees is an hour

Explanation:

What is the time difference in mean solar time between 30° N 75° W and 30° N 90° W?

5 0

2 years ago

key A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration o

lorasvet [3.4K]

Answer:

2.41065 grams

Explanation:

Here we have to apply molarity, particularly in reference to the equation molarity = moles of solute / volume. I would like to rewrite this formula, but with respect to the units - grams = moles / Liters,

We can use molarity to determine the number of moles. After doing so, we can determine the mass of the solute with respect to the formula moles = mass / molar mass. The molar mass of NaCl is 58.44 grams.

_______________________________________________________

275 mL = 0.275 L,

Number of Moles of NaCl = 0.150 * 0.275 = 0.04125 moles,

Mass = 0.04125 * 58.44 = 2.41065 grams,

Solution - Mass of NaCl = 2.41065 grams

Hope that helps!

5 0

2 years ago

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Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and

ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0

2 years ago

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

Mars2501 [29]

Answer:

Qm = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0

3 years ago

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A solution containing a mixture of 0.0351 M potassium chromate ( K 2 CrO 4 ) and 0.0537 M sodium oxalate ( Na 2 C 2 O 4 ) was ti

ki77a [65]

Answer:

Explanation:

As it was stated in the comments by other user, the question is incomplete but luckily it was posted the rest of the question, so, I'm gonna answer it with the data of that. If you have another questions there, please submit it again or put it in the comments.

a) Which barium salt will precipite first?

In order to know this, we need to take a look at the Ksp of both salts. The given Ksp are:

Ksp 1: 2.1x10^-10 (BaCrO4)

Ksp 2: 1.3x10^-6 (BaC2O4)

Now, we can see that Ksp1 < Ksp 2, but what's this Ksp value means? a Ksp value means that an aqueous solution will form a precipite (So the solid formed, it's not soluble in water), As Ksp 1 is a smaller value than Ksp2, means that the concentrations of Ba and CrO4 are too small, and therefore, it takes more time to precipite. Therefore, the BaC2O4 will precipite first.

b) Concentration of Ba2+ present to BaCrO4 precipite

In this, we know that the reaction taking place is the following:

BaCrO4(s) <--------> Ba^2+ + CrO4^2- Ksp = 2.1x10^-10

The expression for Ksp is:

Ksp = [Ba][CrO4]

We know the concentration of CrO4 cause this comes from the K2CrO4 so, replacing here, we solve for Ba:

[Ba^2+] = Ksp / [CrO4^2-]

[Ba^2+] = 2.1x10^-10 / 0.0351

[Ba^2+] = 5.98x10^-9 M

c) Concentration of Ba^2+ to reduce 10% oxalate concentration

The 10% concentration of oxalate is:

[C2O4] = 0.0537 * 0.1 = 0.00537 M

So, we do the same thing we did in part b) but solving for C2O4:

BaC2O4(s) <--------> Ba^2+ + C2O4^2- Ksp = 1.3x10^-6

[Ba^2+] = 1.3x10^-6 / 0.00537

[Ba^2+] = 2.42x10^-4 M

d) ratio of oxalate and chromate when Ba concentration is 0.005 M

In this case, we calculate concentration of CrO4 and C2O4 with the value of Ba and it's respective Ksp, and then, calculate the ratio:

[CrO4^2-] = Ksp1 / [Ba^2+] = 2.1x10^-10 / 0.0050 = 4.2x10^-8 M

[C2O4^2-] = Ksp2 / [Ba^2+] = 1.3x10^-6 / 0.0050 = 2.6x10^-4 M

So the ratio is:

[C2O4^2-] / [CrO4^2-] = 2.6x10^-4 / 4.2x10^-8

[C2O4^2-] / [CrO4^2-] = 6.19x10^3

8 0

2 years ago