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Is water denser than vinegar

Umnica [9.8K]

Vinegar is denser than water

3 0

3 years ago

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Balance the following equation by adding the correct mole numbers. __P4 + __C12 → __PCl3

faltersainse [42]

(Blank) P4 + (6) C12 → (4) PCI3

6 0

3 years ago

First: Calculate the number of moles of Kool-Aid powder needed to make 100mL of a 0.1M solution.

morpeh [17]

For this question, we can use the molarity formula, which is as follows:

$Molarity= \frac{moles}{Liters}$

We know that we want a 0.1M solution in 100 mL, so we can plug those into the equation to find the number of moles required:

$0.1M= \frac{moles}{0.1L}$
$moles=0.01$

So now we know that the number of moles of the Kool-Aid powder is 0.01. We also know that the chemical formula for this powder is $C_{12} H_{22} O_{11}$ .

We can find the molar weight of this powder by taking the atomic weight from the periodic table and multiply it by the number of atoms in the compound. Let's find the molar weight of the powder:

Find the atomic weight of carbon:

$C=12.01 g$

There are 12 C's, so we multiply by 12:

$12.01g*12=144.12g$

Then we have 22 H's, each having an atomic weight of 1.008g:

$1.008g*22=22.18g$

And finally there are 11 O's, each with an atomic weight of 15.99g:

$15.99g*11=175.89g$

Then we add up all of the weights:

$144.12g+22.18g+175.89g=342.19g$ -> So we know that the molar weight of the Kool-Aid powder is 342.19g.

Now we can find the amount of grams that we need for the desired solution as follows:

$\frac{342.19g}{1mol}* \frac{0.01mol}{1}=3.42g$

So now we know that we need 0.01 moles of the powder, and we need 3.42g of the powder to make the desired solution.

7 0

3 years ago

Which of the following is not a major source of freshwater pollution?

morpeh [17]

Answer:

treated sewage is not a major source of freshwater pollution.

3 0

3 years ago

If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone

GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

8 0

3 years ago