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Alla [95]
4 years ago
10

Federico has two samples of pure water—sample X and sample Y. Sample X has a volume of 1 L, and sample Y has a volume of 10 L. H

ow do the boiling points of these two samples compare
Chemistry
1 answer:
Lostsunrise [7]4 years ago
8 0

Answer:

The boiling point of sample X and sample Y are exactly the same.

Explanation:

The difference between sample X and sample Y is that they occupy different volumes. However, they both contain pure water. Remember that pure water has uniform composition irrespective of its volume.

Volume does not affect the boiling point as long as the volume is small enough not to give rise to significant pressure changes in the liquid.

The boiling point of a liquid is the temperature at which the pressure exerted by the surroundings upon a liquid is equaled by the pressure exerted by the vapour of the liquid; under this condition, addition of heat results in the transformation of the liquid into its vapour without raising the temperature.

It can be clearly seen from the above that the volume of a solution of pure water does not affect its boiling point hence sample X and sample Y will have the same boiling point.

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Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting
larisa [96]

Answer: Therefore, the concentration of final solution is 4.0\times 10^{-8}M

Explanation:

According to the neutralization law,

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 2.3\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

2.3\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=0.28\times 10^{-5}M

Therefore, the concentration of diluted solution is 0.28\times 10^{-5}M

2) On further dilution

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.28\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

0.28\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=0.034\times 10^{-5}M

Therefore, the concentration of diluted solution is 0.034\times 10^{-5}M

3) On further dilution

M_1V_1=M_2V_2

where,

M_1 = molarity of stock solution = 0.034\times 10^{-5}M

V_1 = volume of stock solution = 3.00 ml

M_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 25.00 ml

0.034\times 10^{-5}M\times 3.00=M_2\times 25.00

M_2=4.0\times 10^{-8}M

Therefore, the concentration of final solution is 4.0\times 10^{-8}M

5 0
3 years ago
ASAAAP HURRRY PLS!!!
Setler79 [48]
C or D! When reduced coenzymes produce in the process of the cycle, are oxidized in the process of oxidative phosphorylation in electron transport chain! I hope this helps :(
3 0
3 years ago
Read 2 more answers
Choose the correct answer to make the statement true.
yaroslaw [1]
D.
The prefix “exo” indicates a release. “-thermic” indicates heat. Because there is a release of heat, the reaction gives off heat and is warm to the touch. ΔH is negative because there is a loss of heat energy.
3 0
3 years ago
a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

#SPJ4

3 0
1 year ago
How many hydrogen atoms are in 5.80 mol of ammonium sulfide
IRINA_888 [86]

Moles of ammonium sulfide =  5.80 mol

The formula of ammonium sulfide is (NH₄)₂S

So each molecule of ammonium sulfide has (4 x 2)  or 8 atoms of H

One mole of ammonium sulfide has 8 moles of H

5.80 mol of ammonium sulfide has (8 x 5.8) or 46.4 moles of H

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ atoms

46.4 moles of H x (6.022 x 10²³ atoms/ 1 mole of H)

= 2.8 x 10²⁵ H atoms

Therefore, 2.8 x 10²⁵ H atoms are in 5.80 mol of ammonium sulfide.

5 0
3 years ago
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