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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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Some chemical changes can be ______________ through another _______________ change.

zalisa [80]

Some chemical changes can be reactive through another chemical change?

3 0

3 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

3 years ago

The balanced redox reactions for the sequential reduction of vanadium are given below.

Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

$2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)$ $2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)$

$2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)$

On adding above equations, we get the following combined equation.

$2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)$

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

$12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}$

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

$0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn$

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

$0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn$

We need 0.0392 grams of Zn metal to completely reduce vanadium.

6 0

3 years ago

A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A

Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0

3 years ago

WILLL GIVE BRAINLIEST IF YOU ANSWERRR PLEASEEEE IM LITERALLY BEGGING YOU I PUT THIS QUESTION IN SOO MANY TIMES I HAVEN'T GOTTEN

tino4ka555 [31]

Answer:

The answer is 375.54 g of AgBr

Explanation:

Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)

Mass = 2M x 1L x 187.77 g/mol

Mass = 375.54g

6 0

2 years ago