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Darya [45]
3 years ago
7

What layer in the atmosphere has rain clouds

Chemistry
2 answers:
Ilya [14]3 years ago
4 0
Troposphere has the most rain clouds
Hitman42 [59]3 years ago
3 0
Troposphere has most of our clouds temperature etc
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Label the equivalence point on the graph of pH versus volume of the titration of a strong acid and strong base shown below
Lyrx [107]
For C, they would be the same. At the equivalence, you have equal moles of both the acid and the base.
4 0
3 years ago
Can some one please help me with this question with the steps please, thanks in advance ​
tekilochka [14]

The molecular formula for Hydrocarbon =  C₄H₁₀

<h3>Further explanation:</h3>

Given

50 ml hydrocarbon

200 ml CO₂

250 ml H₂O

Required

The molecular formula of Hydrocarbon

Solution

From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles  

So moles Hydrocarbon : CO₂ : H₂O = 50 ml : 200 ml : 250 ml = 1 : 4 : 5

mol C in 1 mol CO₂=1, and for 4 moles CO₂ there are 4 moles C

mol H in 1 mol H₂O =2, and for 5 moles H₂O there are 10 moles H

So mol ratio C : H in compound = C₄H₁₀

4 0
3 years ago
In a laboratory investigation, an HCl(aq) solution with a pH value of 2 is used to determine the molarity of a KOH(aq) solution.
guapka [62]

Answer:

blue

Explanation:

KOH is a base therefore it's pH will be above seven. According to table M when tested with the indicator bromcreeol green The solution will turn blue

5 0
4 years ago
State the Alfabau's principle in atomic chemistry
Jlenok [28]
According to the Aufbau principle, , electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). 
7 0
3 years ago
calculate the freezing point of 3.46 gram of a compound X in 160 gram of benzene when a separate sample of X was vaporized it's
Temka [501]

Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is 4.38^0C

Explanation:

The relation of density and molar mass is:

d=\frac{PM}{RT}

where

d = density = 3.27 g/ L

P = pressure of the gas  = 773 torr = 1.02 atm   (760 torr = 1atm)

M = molar mass of the gas  = ?

T = temperature of the gas = 116^0C=(116+273)K=389K

R = gas constant  = 0.0821Latm/Kmol

M=\frac{dRT}{P}=\frac{3.27g/L\times 0.0821Latm/Kmol\times 389K}{1.02atm}=102.3g/mol

The relation of depression in freezing point with molality:

\Delta T_f=k_f\times m

\Delta T_f = depression in freezing point = T_f^0-T_f = 5.45-T_f

k_f = freezing point constant  = 5.1

m = molality = \frac{\text {moles of X}}{\text {weight of solvent in kg}}=\frac{3.46\times 1000}{102.3\times 160}=0.21

5.45-T_f=5.1\times 0.21

T_f=4.38^0C

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is 4.38^0C

5 0
3 years ago
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