Is a charge indicated on neutral atoms

What is the solution's freezing point: 15 g of CH4N2O (Molar mass = 60.055 g/mol) in 200. g of H2O? (Kf = 1.86 (°C·kg)/mol)

AURORKA [14]

Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :

1 mol of urea =15/60.055 = 0.25mol

therefore 200g of water contain 0.25mol

the next step is to determine the malality of our solution in 200g of water, to do this we say:

200 g = 1Kg/1000g = 0.2kg

therefor 0.25mol/0.2Kg = 1.25mol/kg

and from the equation:

we know that i = 1

we are given Kf

b is the molality that we just calculated

therefore;

the solutions freezing point is -2.325°C

8 0

1 year ago

Check my answer plz!! 100 points. Tell me if they are wrong! 1. You have three elements, A, B, and C, with the following electro

Artyom0805 [142]

look good w small changes below:

Answer:

AB is an ionic compound. The electronegativity difference between A and B is greater.

AC is an ionic compound. The electronegativity difference between A and C is greater.

BC is a covalent compound because the electronegativity difference between C and B is small.

everything else look good!

6 0

3 years ago

Read 2 more answers

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium

babunello [35]

First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume

= 0.15 m * 0.025 L

= 0.00375 moles

moles NaOH = molarity * volume

= 0.15 m * 0.015 L

= 0.00225 moles

after that we shoul get the total volume = 0.025L + 0.015L

= 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume

= 0.00375 moles / 0.04 L

= 0.09375 M

∴[NaOH] = moles NaOH / total volume

= 0.00225 moles / 0.04 L

= 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka

= - ㏒ 1.9 x10^-5

= 4.7

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48

6 0

3 years ago

For the following problems, identify the substance as either an element, a compound, or a mixture. Write E for Element, C for co

andre [41]

Answer:

Explanation:

34. E

35. M

36. C

37. M

38. E

39. C

3 0

3 years ago