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3 years ago

What is the formula for Silver (I) chloride?

Chemistry

2 answers:

n200080 [17]3 years ago

8 0

Answer:

AgCl formula for silver(l) chloride

Kisachek [45]3 years ago

6 0

The formula for Silver (I) chloride is: AgCl

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Consider the titration of sulfuric acid with sodium hydroxide. What volume (mL) of a 2.658M NaOH solution is required to fully t

KatRina [158]

The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

From the question;

Concentration of acid CA = 0.426M

Concentration of base CB = 2.658M

Volume of acid VA = 10.00mL

Volume of base VB = ?

Number of moles of acid NA = 1

Number of moles of base NB = 2

Using the relation;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB = 0.426M × 10.00mL × 2/ 2.658M × 1

VB = 3.2 mL

Learn more: brainly.com/question/6111443

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3 years ago

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victus00 [196]

Answer is d. in hetrogeneous you can separate things from each other

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3 years ago

Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was react

stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

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3 years ago

Which ones are false? need help please.

kirill [66]

Answer:

A.) False B.)False C.) True D.) False E.) False

Explanation:

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3 years ago

Read 2 more answers

A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will b

Angelina_Jolie [31]

Answer:

50.4 % of the water will be vapor

Explanation:

<u>Step 1:</u> Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

<u>Step 2:</u> Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V)/(R*T)

n = (0.2467 * 5.3) /(0.08206* 338)

n = 0.047 moles

<u>Step 3:</u> Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles *18.02 g/mol

Mass of water = 0.84694 grams

<u>Step 4:</u> Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) *100%

% = 50.4%

50.4 % of the water will be vapor

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3 years ago