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Dafna11 [192]
3 years ago
10

What molarity is prepared when a solution contains 0.277 moles of calcium bromide in 500 mL of solution (must change mL to Liter

s)? *
Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0

Answer:

0.554M of Calcium Bromide

Explanation:

Molarity by defintion is #of moles of something/litres of solution.

Therefore, here, we have 0.277 moles of calcium bromide and 500mL (divide 500ml by 1000 to go from mL to L because for every 1L there's 1000mL) or 0.5L.

Molarity= 0.277/0.5 = 0.554M of Calcium Bromide

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3 years ago
The temperature of nitrogen is unknown when the gas occupies 100.0 mL at 99.10 kPa. If the gas is known to occupy 74.2 mL at 133
laiz [17]

Answer:

The unknown temperature is 304.7K

Explanation:

V1 = 100mL = 100*10^-3L

P1 = 99.10kPa = 99.10*10³Pa

V2 = 74.2mL = 74.2*10^-3L

P2 = 133.7kPa = 133.7*10³Pa

T2 = 305K

T1 = ?

From combined gas equation,

(P1 * V1) / T1 = (P2 * V2) / T2

Solving for T1,

T1 = (P1 * V1 * T2) / (P2 * V2)

T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)

T1 = 3022550 / 9920.54

T1 = 304.67K

T1 = 304.7K

3 0
3 years ago
Which gas would act most like an ideal gas? a. chlorine gas at 32oC b. chlorine gas at -185oC
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A as the lower temperature in Celsius corresponds to a lower Kelvin temperature thus reducing movement
6 0
3 years ago
Nitrogen us a component of all​
Ede4ka [16]

Answer:

amino acids and urea

Explanation:

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4 0
3 years ago
A buffer is prepared by adding 139 mL of 0.39 M NH3 to 169 mL of 0.19 M NH4NO3. What is the pH of the final solution? (Assume th
Paha777 [63]

Answer:

pH = 9.48

Explanation:

We have first to realize that NH₃ is a weak base:

NH₃ + H₂O ⇔ NH₄⁺ + OH⁻     Kb = 1.8 x 10⁻⁵

and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.

Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:

pOH = pKb + log ( [ conjugate acid ] / [  weak base ]

mol NH₃ = 0.139 L x 0.39 M = 0.054 mol

mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol

Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)

pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52

pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48

The solution is basic which agrees  with NH₃ being  a weak base.

5 0
3 years ago
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