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den301095 [7]
3 years ago
10

What mass of sucrose c12h22o11 is needed to make 300 ml of a 0.50m solution?

Chemistry
1 answer:
zepelin [54]3 years ago
5 0
Use the concentration to obtain the moles. I am assuming you mean to write capital M. because little m means molality. 

So, first convert the ml into Liters and then into moles, then moles to grams using the molar mass (just adding the values of each atom from the periodic table. )

Molar mass= 12 (12.0) + 22 (1.01)+ 11 (16.0)= 342 grams/mole

300 ml (1 liter/ 1000 mL) x (0.50 moles/ 1 Liter) x (342 grams/ 1 mole)= 51.3 grams


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Explanation:

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3 years ago
Consider the connotation of the underlined word in each sentence, and then match each sentence to the best
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7 0
4 years ago
How many moles of Manganese there in are 5.76 x 10(15) atoms of Mn?
aksik [14]

Answer:

1. 9.57 × 10^-9 moles.

2. 7.38mol

Explanation:

1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)

Hence, to find the number of moles (n) of Manganese (Mn), we say:

5.76 x 10^15 atoms ÷ 6.02 × 10^23

5.76/6.02 × 10^(15-23)

= 0.957 × 10^-8

= 9.57 × 10^-9 moles.

2.) Mole = mass/molar mass

Molar mass of sodium chloride (NaCl) = 23 + 35.5

= 58.5g/mol

mole = 431.6 g ÷ 58.5g/mol

mole = 7.38mol

7 0
3 years ago
Please help me like now please
777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

<u>Question 1</u>

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

<em>To balance the equation, count the number of atoms on both sides of the equation</em>

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

<em>Combining the pluses will give you the following</em>

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

<em>Both sides are the same, therefore the chemical equation is balanced (originally). </em>

From the equation, we can see that <u>1 mole of NaHCO3</u> produces <u>1 mole of CO2</u>.

So that means <u>0.5 mole of NaHCO3</u> would produce <u>0.5 mole of CO2</u>.

<u>Question 2</u>

C4H10 + O2 --> CO2 + H2O

<em>Again, count the number of atoms on both sides of the equation</em>

(4C, 10H, 2O) --> (1C, 2H, 3O)     <em>This time left does not equal right side</em>

<em>You now need to find </em><u><em>factors </em></u><em>that can make both sides equal. </em>

C4H10 + O2 --> <u>4</u>CO2 + H2O    <em>Now the C is balanced, let's recount </em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H)      H&O is still not balanced</em>

C4H10 + O2 --> 4CO2 + <u>5</u>H2O    <em>Now the H is balanced, let's recount</em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H)      O is still not balanced</em>

C4H10 + (<u>13/2</u>)O2 --> 4CO2 + 5H2O    <em>Now the O is balanced</em>

<em>(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)</em>

<em>But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).</em>

<u>2</u>C4H10 + <u>13</u>O2 --> <u>8</u>CO2 + <u>10</u>H2O    Now it's completely balanced!

<em>(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H)     Yayy! It's balanced.</em>

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0
3 years ago
What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?
Viefleur [7K]
Data:
Molar Mass of NaOH = 40 g/mol

Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³ 
y =  \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}
\boxed{y \approx 0.04\:mol}


Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

x = 40*0.04
\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark

Answer:
The mass is 1.6 grams
6 0
3 years ago
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