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3 years ago

PLEASE HELP ME.

Chemistry

2 answers:

vladimir2022 [97]3 years ago

8 0

I think the answer is b

Andrej [43]3 years ago

7 0

Velocity and speed both take into account the change is distance over time. But velocity also considers a body’s change in position—that is, velocity accounts for directionality, while speed does not. So, the correct answer here would be C.

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What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (Kf = 1.86°C/m) –1.86°C –7.44°C –5.58°C –3.72°C

tresset_1 [31]

Answer:

Last choice: <em><u>- 3.72°C</u></em>

Explanation:

The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.

The equation to predict the freezing point depression in a solvent is:

ΔTf = Kf × m × i

Where,

ΔTf is the freezing point depression of the solvent,

m is the molality,

Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.

The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.

Download pdf

6 0

2 years ago

Sarah rolls a ball in the grass. The ball starts off moving quickly, starts to slow down, and eventually stops. Using the words

gogolik [260]

Answer:

hi 5th grader, stop trying to cheat :)

Explanation:

4 0

2 years ago

Which situation would cause the following equilibrium reaction to decrease the formation of the products? 2SO2 (g) + O2 (g) Two

vovangra [49]

The given equilibrium reaction is,

$2 SO_{2} (g) + O_{2}(g) 2 SO_{3} (g) + Energy$

The given reaction is exothermic. So, heat energy will be a product. Therefore, decreasing the temperature (heat energy) would lead to the formation of more products as when the amount of energy which is a product is reduced, there is more room for the products to form.

Increasing the pressure would shift the equilibrium towards that side which has least number of moles of the gaseous substance. Hence, here increasing the pressure would lead to the formation of more products by shifting the equilibrium towards the right side.

Decreasing the volume would make the equilibrium shift towards the least number of moles of the gaseous substance. So, here in this equilibrium decreasing the volume would lead to the formation of more products.

8 0

2 years ago

Read 2 more answers

I WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST

snow_tiger [21]

Answer:

I think it's B

Explanation:

I might be wrong, but please tell me.

5 0

2 years ago

DUE TOMORROW!!! 15 POINTS

Lunna [17]

Answer:

A. write balanced chemical equation (including states), for this process.

Explanation:

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve

[some hydrocarbon] + oxygen --> carbon dioxide + steam.

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side:

C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side:

C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left.

Two ways to deal with that. We can use a fraction:

C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything

to get rid of the fraction:

2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in...

C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

6 0

3 years ago