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3 years ago

QUICK! WILL MARK BRAINLIEST

Chemistry

1 answer:

nalin [4]3 years ago

5 0

Answer:

Ability to be bent = Malleability

Identity = Physical Change

Electrical Current = Conductivity

Dissolve = Solubility

Color, Phase, or Hardness = Physical Property

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In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver

pantera1 [17]

Answer:

The mass in grams of silver chloride, AgCl produced is 84.37 grams

Explanation:

The chemical equation for the reaction is given as follows;

AgNO₃ + NaCl → AgCl + NaNO₃

The mass of silver nitrate, AgNO₃, in the reaction = 100.0 g

Therefore, 1 Mole of AgNO₃ produces 1 mole of AgCl,

The molar mass of silver nitrate, AgNO₃ = 169.87 g/mol

The number of moles of silver nitrate, AgNO₃, in the reaction = 100/169.87 moles

The number of moles of silver nitrate, AgNO₃, in the reaction ≈ 0.59 moles

Since 1 mole of AgNO₃ produces 1 mole of AgCl, 0.59 moles of AgNO₃ produces 0.59 moles of AgCl

The number of moles of silver chloride, AgCl produced = 0.59 moles

The molar mass of silver chloride, AgCl = 143.32 g/mol

Therefore;

The mass of silver chloride, AgCl produced = The number of moles of silver chloride, AgCl produced × The molar mass of silver chloride, AgCl

Which gives;

The mass of silver chloride, AgCl produced = 143.32 g/mol × 0.59 moles = 84.37 g

The mass of silver chloride, AgCl produced = 84.37 g.

7 0

3 years ago

why would air moving over a cold current cause fog (advection fog)? group of answer choices the cold current produces the fog wh

frutty [35]

The correct answer is option C.

The air moving over a cold current cause fog or advection fog when the air is chilled, which decreases the capacity of air to hold water vapor, and eventually, the relative humidity reaches 100% - leading to condensation and fog formation.

What is advection fog?

When warm- moist air or warm air front slides over the cold air front or cold surface, it results in the formation of advection fog.

Resultantly, the air becomes saturated and chilled at high humidity levels due to which water vapors start to condense leading to fog formation.

Moreover, the optimal condition for the formation of advection fog is cloudy windy weather having moderate to powerful winds.

If you want to learn more about advection fog click here:

brainly.com/question/18803983

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The complete question is:

Why would air move over a cold current cause fog (advection fog)?

(a) the cold current produces the fog when kelp beds release condensation nuclei.

(b) the cold current produces the fog by mixing with the air.

(c) the air is chilled, which decreases the capacity of air to hold water vapor, and eventually, the relative humidity reaches 100% - leading to condensation and fog formation.

7 0

1 year ago

Please help I just need those last 4! :)

mojhsa [17]

Answer:

Series:

3. Thermistor

Parallel:

1. ???

2. ???

3. ???

5 0

3 years ago

For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.

Iteru [2.4K]

a) $2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

b) $2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

$2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

$E^0 cell = E^0 (reduction) - E^0 (oxidation)$

= $E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)$

= -0.83 - (-2.71) =1.88V

$2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

$E^0cell$ = $E^0(H^+/H_2) -E^0(Ag^+/Ag)$

$E^0cell$ =-0. - (0.8) =-0.8V

Learn more about the half-reactions here:

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8 0

2 years ago

Calculate the value of the equilibrium constant, Kc , for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2,

sdas [7]

the reaction is

2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

Given

moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062

moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54

4 0

2 years ago