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Which of the following is not true? a. The freezing point of sea water is lower than the freezing point of pure water. b. The bo

Anna007 [38]

Answer:

The false statement is b.

The boiling point of a water sample from the Salt Lake is lower than the boiling point of pure water

Explanation:

This excersise refers to colligative properties.

Boiling point refers to the property of boiling point elevation. In a solution of ions, as water sea or water from lakes, the boiling point will be higher than water pure.

It is logical to say that, because these sort of water have ions. Remember that colligative properties depends on the solute particles.

T° boiling solution - T° pure solvent = Kb . m . i

In the freezing point depression, we have the oppossite of boiling point elevation. Freezing point of solution is lower than pure solvent, according to this:

T° freezing pure solvent - T° freezing solution = Kc . m . i

We have to always consider the i, which means Van't Hoff factor, number of ions dissolved in solution. As the i is higher, the freezing point of solution will be lower, and the boiling point of solution will be higher.

They are true statement a and c.

In solution of., [NaCl] = 0.1M, as it is a higher concentration, the molality is also higher than a solution of [NaCl] = 0.05.

As water sample form Salt Lake has certain ions, it is logical to say that the boiling point of this water is higher than boiling point of pure water.

5 0

3 years ago

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0

3 years ago

Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

Whitepunk [10]

To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020

Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹

7 0

3 years ago

What reactions are responsible for the glow and heat from the sun? nuclear fission nuclear fusion chemical reactions atomic disi

Mamont248 [21]

Nuclear Fusion is the answer to the question who posted.

7 0

3 years ago

Read 2 more answers

If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? A. 196 N B. 32.67 N End of exam C. 1.96 kg

garri49 [273]

Force = mass x gravity

Force = 20 kg x 9.8 m/s²

Force = 196 Newtons

Answer A

hope this helps!

7 0