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3 years ago

CAN SOMEONE HELP ME PLEASE . i have the definitions i just need help with examples

Chemistry

1 answer:

Alla [95]3 years ago

6 0

Collision Coverage Example:
You're driving down the street and rear end the car in front of you. Your liability insurance will pau for the damage done to the other car, and your Collision insurance (or coverage) will pay for the damage done to your own vehicle because this was an accident in which you collided with another vehicle.

Comprehensive Coverage Example:
When there's damage to your windshield; damage from natural disasters and bad weather conditions, like wind storms, hurricanes, floods, earthquakes, hail, etc,.

Hope I helped!

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Question 1 (1 point)

stealth61 [152]

Answer:

Q1: 9

Q2:10

Q3: 50

Q4:True

Q5: False

Q6:False

Q7:True

Explanation: Just took the quiz

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3 years ago

Mars is known as the ____ planet. A) Blue B) Gaseous C) Red D) Sandy

Leni [432]

Answer:

Of course it's C

Red planet

Explanation:

It is because the soil on Mars is rich of Fe (Iron).

That makes the soil look red.

Even on our planet we have such this places like hormuz island in Iran.

5 0

3 years ago

In this experiment you will create solutions with different ratios of ethanol and water. What is the mole fraction of ethanol wh

Andrej [43]

Answer:

The mole fraction of ethanol is 0.6. A 10 mL volumetric pipette must be used for to measure the 10 mL of ethanol. The vessel should be clean and purged.

Explanation:

For calculating mole fraction of ethanol, the amount of moles ethanol must be calculated. Using ethanol density (0.778 g/mL), 10 mL of ethanol equals to 7.89 g of ethanol and in turn 0.17 moles of ethanol. The same way for calculate the amount of water moles (ethanol density=0.997 g/mL). 2 mL of water correspond to 0.11. The total moles are: 0.17+0.11=0.28. Mole fraction alcohol is: 0.17/0.28=0.6

5 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers

You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye

dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

$V$ = 10 mL solution

$V_{added}$ = 12.20 mL

$V_{total}$ = 22.20 mL

then,

M 0.103 M of NaOH,

$V_{rinsed}$ = experiment will not be affected

$V_{total \ base}$ = 10.38 mL

Now,

⇒ mol of NAOH = MV

= $0.103\times 10.38$

= $1.06914 \ m$

Whether Sulfuric acid, then

⇒ $H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O$

⇒ $mol \ of \ acid =\frac{1}{2}\times \ mol \ of \ base$

⇒ $1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid$

Before any dilution:

$V_{sample} = 10 \ mL$

⇒ $M \ acid = \frac{m \ mol}{V}$

$=\frac{ 0.53457 }{10}$

$=0.053457 \ M$ (Sulfuric acid)

6 0

3 years ago