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3 years ago

How many milliliters of a 10.6 M NaOH solution are needed to prepare 0.715 L solution of NaOH with a concentration of 0.550 M?

Chemistry

1 answer:

dusya [7]3 years ago

7 0

Answer:

37.1mL

Explanation:

This is a simple dilution problem where you utilize the equation:

(C=Concentration, V=Volume)

C1V1=C2V2

10.6(y)=0.715x0.550 (Y is unknown because we are trying to find how much of the original solution we need to dilute to make the final solution)

The solve for y and you get

y= 0.0317 L but this is in L and we want mL so multiply by 1000 you get

y= 37.1mL

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What is the total mass of three gold bars that weigh 5543 mg, 23.45 mg, and 697.4 mg?

GrogVix [38]

The total mass is the sum of the masses.
It is
5543 + 23.45 + 697.4 mg = 6263.85 mg

Answer: 6263.85 mg

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2 years ago

What volume of 0.555M KNO3 solution would contain 12.5 g of solute

Lorico [155]

The volume of 0.555M KNO3 solution would contain 12.5 g of solute iss 223 mL.

<h3>What is the relationship between mass of solute and concentration of solution?</h3>

The mass of solute in a given volume of solution is related by the formula below:

Molarity = mass/(molar mass * volume)

Therefore, volume of solution is given by:

Volume = Mass /molarity * molar mass

Molar mass of KNO₃ = 101 g/mol

Volume = 12.5/(0.555 * 101)

Volume = 0.223 L or 223 mL

In conclusion, the volume of the solution is obtained from the molarity of solution as well as mass and molar mass of solute.

Learn more about molarity and volume at: brainly.com/question/26873446
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3 0

1 year ago

How many grams of oxygen are produced when 1.05 moles of hydrogen gas is<br> produced?

vivado [14]

Answer:

1.058337 grams of hydrogen and 2H2 + O2 ==> 2H2O hydrogen peroxide

mols H2 = grams/molar mass

Using the coefficients in the balanced equation, convert mols H2 to mols O2.

Now convert mols O2 to grams. That's grams mols O2 x molar mass O2.

and it could also produce H2O water but no air but it could make other things

8 0

2 years ago

What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?

Step2247 [10]

Answer:

$\Delta \rho =2.22 g/mL$

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

$\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL$

Best regards.

3 0

2 years ago

Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1

tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0

2 years ago