1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vlabodo [156]
2 years ago
6

Map

Chemistry
1 answer:
grin007 [14]2 years ago
3 0

Answer: 1650 hope i got it in time.

Explanation:

You might be interested in
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
Please answer a b c :)​
olganol [36]

Answer:

i think the letter b is the answer

Explanation:

sana makatulong

8 0
3 years ago
What is the approximate bond angle around the central carbon atom in acrolein?
Soloha48 [4]

Answer:

The approximate bond angle around the central carbon atom in acrolein is 120°.

Explanation:

The structure of acrolein is shown in the attachment. From the structure, we can deduce that the central carbon atom is in an sp2 hybridization (Atoms with a double bond hybridize in an sp2 fashion).

Atoms with sp2 hybridization have trigonal planar geometry, in this kind of hybridization, bonds are oriented the farthest away possible from each other, to minimize overlapping and the angle that allows that is 120°. 

7 0
4 years ago
What effect does increasing the concentration of a dissolved solute have on each of the colligative properties?
Igoryamba

Answer:

If we increase the concentration of a dissolved solute, the solution would have a vapor pressure so much low, the boiling temperature for the solution will be so high, freezing point for the solution will be so much low and the osmotic pressure will be higher.

Colligative properties always depends on dissolved particles (solute)

Explanation:

These are the colligative properties

- Vapor pressure lowering

ΔP = P° . Xm

Vapor pressure of pure solvent - Vapor pressure of solution.

If we add more solute, it would raise the Xm, so the solution would have a vapor pressure so much low.

Vapor pressure pure solvent - Vapor pressure solution ↑ = P° . Xm ↑

- Boiling point elevation

ΔT = Kb . m

When we add more solute, we are increasing the molality.

↑T° boiling of solution - T° boiling pure solvent = Kf . m ↑

Boiling temperature for the solution will be so high.

- Freezing point depression

When we add more solute, we are increasing the molality.

ΔT = Kf . m

T° fussion of pure solvent - ↓T° fussion of solution = Kf . m↑

Freezing point for the solution will be so much low.

- Osmotic pressure

π = M . R . T

When we add solute, molarity is increasing. Therefore the osmotic pressure will be higher.

π↑ = M↑ . R . T

7 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
3 years ago
Other questions:
  • What type of bond would you expect in a compound of oxygen & sodium?
    8·1 answer
  • This is a Science Question:
    14·1 answer
  • Products that are poisonous to you and pets are known as: A) flammables. B) reactives. C) toxins. D) corrosives
    11·1 answer
  • What is the answer: 20°C to °F?
    10·1 answer
  • 14 POINTS
    10·2 answers
  • Find the effective resistance of the circuit
    15·1 answer
  • What effect does super critical mass have on a nuclear reaction The reaction rate increases b The reaction produces less energy
    6·2 answers
  • 26. At high temperatures these elements conduct electricity but at lower temperatures they do not conduct electricity. This make
    11·2 answers
  • The drawing below was sketched by a student who was observing cells under a microscope. The cell is classified as
    8·1 answer
  • (06.04 HC)
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!