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Lilit [14]
3 years ago
14

A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask

to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.
Chemistry
1 answer:
leonid [27]3 years ago
8 0

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer:  The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

V_s = volume of solution in ml

moles of Ba(CH_3COO)_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol

Now put all the given values in the formula of molality, we get

Molarity=\frac{0.125\times 1000}{350ml}

Molarity=0.357M

Therefore, the concentration of solution is 0.375 mol/L

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A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titra
Eddi Din [679]

Answer:

See explanation below

Explanation:

In order to calculate this, we need to use the following expression to get the concentration of the base:

MaVa = MbVb (1)

We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:

Atomic weights of the elements to be used:

K = 39.0983 g/mol;  H = 1.0078 g/mol;  C = 12.0107 g/mol;  O = 15.999 g/mol

MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol

Now, let's calculate the mole of KHP:

moles = 0.5053 / 204.2189 = 0.00247 moles

With the moles, we also know that:

n = M*V (2)

Replacing in (1):

n = MbVb

Now, solving for Mb:

Mb = n/Vb  (3)

Finally, replacing the data:

Mb = 0.00247 / (13.4473/1000)

Mb = 0.184 M

This would be the concentration of NaOH

8 0
3 years ago
199.5 grams unrefined dark crystalline sugar to cups
Dmitry [639]

Answer:

0.9975 cup  

Step-by-step explanation:

"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."

200.0 g brown sugar = 1 cup

 199.5 g brown sugar = 199.5× 1/200 .0

 199.5 g brown sugar = 0.9975 cup

A standard measuring cup is not capable of this precision and, furthermore, the mass of brown sugar you can get into a cup depends on how tightly you pack it.

Your Mole Day cake will be fine if you use 1 cup of brown sugar as usual.

4 0
3 years ago
I REALLY NEED HELP ITS DUE AT 11:59 PM
Slav-nsk [51]

Answer:

I attached a photo of balanced equations but thats as much as I can help.

Explanation:

5 0
3 years ago
Read 2 more answers
What is the molarity of a solution composed of 5.85 g of potassium iodide, KI, dissolved
Troyanec [42]

Answer:

0.282 M

General Formulas and Concepts:

<u>Chemistry - Solutions</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Molarity = moles of solute / liters of solution

Explanation:

<u>Step 1: Define</u>

5.85 g KI

0.125 L

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of I - 126.90 g/mol

Molar Mass of KI - 39.10 + 126.90 = 166 g/mol

<u>Step 3: Convert</u>

<u />5.85 \ g \ KI(\frac{1 \ mol \ KI}{166 \ g \ KI} ) = 0.035241 mol KI

<u>Step 4: Find Molarity</u>

M = 0.035241 mol KI / 0.125 L

M = 0.281928

<u>Step 5: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.281928 M ≈ 0.282 M

7 0
3 years ago
Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um
Andre45 [30]

Answer:

\large \boxed{\text{0.0038 mol/L}}

Explanation:

1. Calculate the initial moles of acid and base

\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}

2. Calculate the moles remaining after the reaction

                   OH⁻     +     H₃O⁺ ⟶ 2H₂O

I/mol:      0.0053       0.005 00

C/mol:    -0.00500   -0.005 00

E/mol:      0.0003              0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}

8 0
3 years ago
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