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STALIN [3.7K]
2 years ago
10

How much NaCl would you dissolve and dilute to the 100.00 mL mark in a volumetric flask with DI H2O to prepare a 0.825 M sodium

chloride solution?
Chemistry
1 answer:
adoni [48]2 years ago
4 0

Answer:

4.82 g

Explanation:

To solve this problem we'll use the <em>definition of molarity</em>:

  • Molarity = moles / liters

We are given the volume and concentration (keep in mind that 100mL=0.100L):

  • 0.825 M = moles / 0.100 L

Now we <u>calculate the number of NaCl moles required</u>:

  • moles = 0.0825 mol

Then we convert 0.0825 NaCl moles into grams, using its molar mass:

  • 0.0825 mol * 58.44 g/mol = 4.82 g
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Gold is alloyed with other metals to increase its hardness in making jewelery.a) Consider a piece of gold jewelry the weighs 9.8
irinina [24]

Answer:

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

Explanation:

Mass of jewellery = 9.85g Volume of jewellery = 0.675cm^3.

Density of gold = 19.3g/cm^3 and density of silver = 10.5g/cm^3

Let the volume of gold in the jewellery be X and the volume of silver in the jewellery be Y

Hence we have

Density = mass/volume or mass = volume × density = for gold = X × 19.3g/cm^3 and for the silver = Y × 10.5g/cm^3

19.3X +10.5Y = 9.85g

Also volume of jewellery is given by

Volume of silver in the jewellery + volume of gold in the jewellery = 0.675cm^3.

X + Y = 0.675cm^3.

Solving the above equations we have

Y = 0.675 - X

Which gives

19.3X + 10.5Y = 9.85g

19.3X + 10.5 × (0.675 - X) =9.85g

19.3X + 7.0875 - 10.5X = 9.85

8.8X + 7.0875 = 9.85

8.8X = 2.7625

or X = 0.3139 cm^3

But Y = 0.675 - X

Hence Y = 0.675 - 0.3139 = 0.3611 cm^3

Mass of gold in the jewellery = volume of gold × Density of gold = 0.3139 × 19.3 = 6.059 g

Also mass of silver = 10.5 × 0.3611 = 3.7913g

A). Percentage of gold by mass of the jewellery =

(6.059÷9.85) × 100 = 61.5%

B). Purity of the gold = 0.615 × 24 = 14.76 karat ~ 15 karat gold

7 0
3 years ago
What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl
Kitty [74]

Answer:

\large \boxed{\text{0.012 mol}}  

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

               CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL:                    70.

c/mol·L⁻¹:             0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }

3 0
2 years ago
What is a good indicator for titrating potassium hydroxide with hydrobromic acid?
Cerrena [4.2K]
Potassium hydroxide is a strong base and hydrobromic acid is a strong acid. This implies that the pH of the end-point [neutralization] of their titration will be around pH 7. A good indicator for this kind of pH is bromthymol blue. This is because this indicator changes its colour at pH 7.
4 0
3 years ago
Calculate E∘cell for the following balanced redox reaction:<br><br> 2Cu(s)+Ca2+(aq)→2Cu+(aq)+Ca(s)
Neko [114]

Answer:

FOK I don’t know. I’ll come back to the question

Explanation:

6 0
3 years ago
You are on the Titanic and want to find something that will float because you didn't
damaskus [11]

Find volume of pillow

L=78cm

B=55cm

H=25cm

\\ \bull\tt\longrightarrow V=LBH

\\ \bull\tt\longrightarrow V=25(78)(55)

\\ \bull\tt\longrightarrow V=107250cm^3

\\ \bull\tt\longrightarrow V=10.72m^3

Now

Mass=5.5kg

\\ \bull\tt\longrightarrow Density=\dfrac{Mass}{Volume}

\\ \bull\tt\longrightarrow Density=\dfrac{5.5}{10.72}

\\ \bull\tt\longrightarrow Density=0.5kg/m^3

Density of water=1000kg/m^3

As it is less than density of water it will float on water

8 0
2 years ago
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