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deff fn [24]
3 years ago
12

How would you calculate the molar mass of 2AgCl?

Chemistry
1 answer:
Rashid [163]3 years ago
3 0
2 × (atomic mass of Ag) + (atomic mass of Cl (
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The balanced reaction equation for combustion of heptane, C7H16, is
Allisa [31]

Answer: m = 11.2 g C7H16

Explanation: First convert the mass of CO2 to moles. Then do the mole ratio between CO2 and C7H16 which is 7:1. Finally convert the moles of C7H16 to the mass of C7H16.

Solution attached.

3 0
3 years ago
Read 2 more answers
6. A 25.0-mL sample of potassium chloride solution was found to have a mass of 25.225 g.
Morgarella [4.7K]

Answer:

6. a. 5.53 (m/m) %; b. 0.7490M

7. 0.156M

Explanation:

6. In the solution of KCl, there are 1.396g of KCl in 25.225g of solution. As mass/mass percent is:

Mass solute / Mass solution * 100

The mass/mass percent of KCl is:

a. 1.396g KCl / 25.225g solution * 100

5.53 (m/m) %

b. Molarity is moles of solute per liters of solution:

<em>Moles KCl -Molar mass: 74.55g/mol-:</em>

1.396g KCl * (1mol / 74.55g) = 0.018726 moles KCl

<em>Volume in liters: 25mL = 0.025L</em>

Molarity:

0.018726 moles KCl / 0.025L = 0.7490M

7. 0.90% means 0.90g of NaCl in 100g of solution:

<em>Moles NaCl -Molar mass: 58.44g/mol-:</em>

0.90g NaCl * (1mol / 58.44g) = 0.0154 moles NaCl

<em>Volume in Liters:</em>

100g * (1mL / 1.01g) = 99mL = 0.099L

Molarity:

0.0154 moles NaCl / 0.099L =

0.156M

4 0
3 years ago
So before i have to do the science fair i have to write 5 questions that will lead me to a hypothesis my project is which substa
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Answer:

Explanation:

The best bet would just be to use filler questions, after all this is one of the least important parts.

Examples would be: Would blank melt ice faster because blank?

and so on

4 0
3 years ago
A blank and blank are examples of solids
galina1969 [7]
Dirt and a chair are examples if solids. Some more examples are tables chips and blankets.
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3 years ago
A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so
uranmaximum [27]

Answer:

<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>

Explanation:

[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>

Ionic equation:

{ \bf{2OH { }^{ - }  _{(aq)} + 2H { }^{ + } _{(aq)}→  2H _{2} O _{(l)} }}

first, we get moles of potassium hydroxide in 28.94 ml :

{ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

{ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\  = { \underline{423.95 \: g \approx424 \: grams}}

for the molar mass:

8 0
2 years ago
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