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marishachu [46]
2 years ago
11

Assume that there is a puddle that contains 12.0 kg of water and it froze outside since it is -1°C. The

Chemistry
1 answer:
Musya8 [376]2 years ago
4 0

The temperature that must be to freeze the solution would be -21.1 ° C.

<h3>How to calculate the freezing temperature of this solution?</h3>

To calculate the freezing temperature we must take into account the following information.

  • Solution with a salt concentration of 10% is frozen at -6°C
  • Solution with a salt concentration of 20% is frozen at -16°C
  • Solution with a higher concentration is frozen at -21.1°C

According to the above, it can be inferred that the puddle has a 50% concentration of salt because they had 12 kg of water and 6 kg of salt.

So the lowest freezing temperature would be 21.1°C because the puddle is 50% concentrated.

Note: This question is incomplete because there is some missing information. Here is the missing information:

  • A 10% salt solution freezes at about 20°F (-6°C), and a 20% solution freezes at 2°F (-16°C).
  • The lowest freezing point obtainable for salt solutions is −21.1 °C

Learn more about freezing in: brainly.com/question/14131507

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Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three s
stiks02 [169]

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x =  1.053 g / 0.0405  mol = 26 g/mol

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4 years ago
What does the word "kinetic" indicate about
Natasha2012 [34]
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2 years ago
I need HELP! What is a variable that meteorologist track to make weather forecast? The answers are either: air pressure or barom
larisa [96]
A barometer is a tool, not a variable. A barometer measures barometric or air pressure, which is a variable that meteorologists use to predict the weather.
3 0
3 years ago
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Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M
GarryVolchara [31]

Answer:

Explanation:

CH₃Br+NaOH⟶CH₃OH+NaBr

It is a single step bimolecular reaction so  order of reaction is 2 , one for CH₃Br and one for NaOH .

rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x .12 x .12

k = .55555

when concentration of CH₃Br is doubled

rate of reaction = .555555 x [.24] [ .12 ]

= .016 M/s

when concentration of NaOH  is halved

rate of reaction = .555555 x [.12] [ .06 ]

= .004  M/s

when concentration of both CH₃Br and Na OH is made 5 times

rate of reaction = .555555 x .6 x .6

= 0.2 M/s

8 0
3 years ago
A student needs to prepare 50.0 mL of a 1.20 M aqueous H2O2 solution. Calculate the volume of 4.9 M H2O2 stock solution that sho
Nonamiya [84]

Answer : The volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

Solution : Given,

Molarity of aqueous H_2O_2 solution = 1.20 M = 1.20 mole/L

Volume of aqueous H_2O_2 solution = 50.0 ml = 0.05 L

(1 L = 1000 ml)

Molarity of H_2O_2 stock solution = 4.9 M = 4.9 mole/L

Formula used :

M_1V_1=M_2V_2

where,

M_1 = Molarity of aqueous H_2O_2 solution

M_2 = Molarity of H_2O_2 stock solution

V_1 = Volume of aqueous H_2O_2 solution

V_2 = Volume of H_2O_2 stock solution

Now put all the given values in this formula, we get the volume of H_2O_2 stock solution.

(1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2

By rearranging the term, we get

V_2=0.01224L=12.24ml

Therefore, the volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

3 0
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