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Ksenya-84 [330]
3 years ago
5

Victor pours 100 grams of oil into a beaker filled with 1,000 grams of water. What is the total mass of his mixture? A. 100,000

grams B. 1,100 grams C. 900 grams D. 10 grams
Chemistry
2 answers:
Anna71 [15]3 years ago
7 0

The answer is 1,100 Grams.

tino4ka555 [31]3 years ago
6 0
The total mass of his mixture is 100,000 grams. (Answer A)
You multiply 1,000 grams of water by 100 to get 100,000 grams.
So the total mass of Victor's mixture is 100,000 grams
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steposvetlana [31]

In general, salts (formed during a neutralization reaction) are ionic compounds that are soluble in water and dissociate in solution into ions that conduct electricity. Out of the six statements given, there are three related statements that rehash the foregoing, and there are three related statements that are collectively incorrect.

Statements A, B, and D are (generally) true regarding salts formed during a neutralization reaction. When you consider that the net ionic equation of many acid-base neutralization reactions is H⁺(aq) + OH⁻(aq) → H₂O(l), the counterions of the H⁺(aq) and OH⁻(aq) are the aqueous spectator ions that comprise the salt. These ions are electrolytes, as they are charged species that can carry a current in solution; they are ionic compounds by definition since they're composed of cations and anions; and, as aqueous species, they're clearly dissolved in water.

Statements C, E, and F, as a whole, generally aren't true of such salts.

4 0
3 years ago
Hikers have noticed that a sealed bag of potato chips puffs up when taken to the top of a mountain. Suppose that at the valley f
Dmitriy789 [7]
<h3>Answer:</h3>

0.95 atm

<h3>Explanation:</h3>

We are given;

Initial pressure, P1 = 1.0 atm

Initial temperature, T1 =298 K (25°C + 273)

Initial volume, V1 = 0.985 L

Final temperature, T2 = 295 K (22°C + 273)

Final volume, V2 = 1.030 L

We are required to find final air pressure;

Using the combined gas law;

\frac{P1V1}{T1} =\frac{P2V2}{T2}

To get, P2 ;

P2=\frac{P1V1T2}{T1V2}

P2=\frac{(1.0)(0.985L)(295)}{(1.030)(298)}

P2 = 0.947 atm

        = 0.95 atm

Therefore, the air pressure at the top of the mountain is 0.95 atm

3 0
3 years ago
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Darya [45]
The answer is A my good sir
7 0
3 years ago
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In 3.00 x 10^20 molecules of C12h22O11, how many C atoms are presented?
tigry1 [53]

Answer:

3.6 x 10²¹ carbon atoms

Explanation:

Data Given:

C₁₂H₂₂O₁₁ = 3.00 x 10²⁰ molecules

Carbon atoms = ?

Solution:

Step 1.

First find number of moles of C₁₂H₂₂O₁₁

Formula used

             no. of moles = no. of molecules/ Avogadro's number

Put vales in above formula

             no. of moles =  3.00 x 10²⁰ / 6.23 x 10²³

             no. of moles =  4.82 x 10⁻⁴ mol

Step 2.

Now find mass of 4.82 x 10⁻⁴ moles of C₁₂H₂₂O₁₁  

Molar mass C₁₂H₂₂O₁₁ = 12(12) + 22(1) + 11(16)

Molar mass C₁₂H₂₂O₁₁ = 342 g/mol

Formula used

                no. of moles = mass in grams / Molar mass

Put values in formula

               4.82 x 10⁻⁴ mol =  mass in grams / 342 g/mol

Rearrange the above equation

              mass in grams =   4.82 x 10⁻⁴ mol x 342 g/mol

              mass in grams =   0.165 g

So,

C₁₂H₂₂O₁₁ = 0.165 g

Step 3.

calculate the percent composition of Carbon (C) in C₁₂H₂₂O₁₁

Since the percentage of compound is 100

So,

Formula used:

Percent Composition of Carbon (C) = mass of carbon  / molar mass x 100

Put values in formula

Percent Composition of Carbon (C) = 144  / 342 x 100

Percent of Carbon (C) = 42 %

It means that for ever gram of C₁₂H₂₂O₁₁ there is 0.42 g of C is present.

So,

For the 0.165 g of C₁₂H₂₂O₁₁ the mass of C will be

mass of Carbon (C) = 0.42 x 0.165 g

mass of Carbon (C) = 0.0693 g

Step 5.

Now find the number of moles for 0.0693 g of Carbon (C)

Molar mass of C = 12 g/mol

Formula Used

            no. of moles = mass in grams / Molar mass

put values in above formula

              no. of moles = 0.0693 g  / 12 g/mol

               no. of mol = 0.0058

no of moles of Carbon = 0.0058

Step 5.

Now find number of atoms in 0.0058 moles of carbon

Formula used

             no. of moles = no. of atoms of C / Avogadro's number

Put vales in above formula

             0.0058 = no. of atoms of C / 6.23 x 10²³

Rearrange the above equation

             no. of atoms of C =  0.0058 x 6.23 x 10²³

             no. of atoms of C =  3.6 x 10²¹

So,

3.6 x 10²¹ carbon atoms are present in 3.00 x 10²⁰ molecules of C₁₂H₂₂O₁₁

6 0
3 years ago
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