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frutty [35]
4 years ago
9

Give the formula for molarity

Chemistry
1 answer:
Scilla [17]4 years ago
4 0

Divide the number of moles by the number of liters

Write your answer.

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The equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of ox
Vaselesa [24]
<span>0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles</span>
7 0
3 years ago
Pls help question and answers in picture <br> Will make you brainliest <br> TYYYY
Burka [1]

Answer:

g

Explanation:

g

7 0
3 years ago
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What are interactions between flip flops and sulfuric acid
frosja888 [35]
When the flip flops are used to walk on the ground the sulfuric acid from the ground overtime tends to wear down the bottom of the flip flops
3 0
3 years ago
The mass of an average blueberry is 0.75 g and the of am automoblie is 2.0 x 10 to the 3 Kg. Find the number of automobiles whos
Paul [167]

Answer : The number of automobiles needed are, 2.26\times 10^{17}

Explanation :

First we have to calculate the mass of blueberries.

As, 1.0 mole of blueberries contains 6.022\times 10^{23} blueberries

So, 0.75 grams of blueberries contains 0.75g\times 6.022\times 10^{23}=4.52\times 10^{23}g blueberries

Mass of blueberries = 4.52\times 10^{23}g

Now we have to calculate the number of automobiles needed.

Number of automobiles needed = \frac{4.52\times 10^{23}g}{2.0\times 10^{3}kg}

Number of automobiles needed = \frac{4.52\times 10^{23}g}{2.0\times 10^{3}\times 10^{3}g}

Number of automobiles needed = 2.26\times 10^{17}

Thus, the number of automobiles needed are, 2.26\times 10^{17}

8 0
3 years ago
Determine the number of ions produced in the dissociation of the compound listed. BaS
11Alexandr11 [23.1K]
BaS → Ba²⁺ + S²⁻
two types of ions
4 0
4 years ago
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