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3 years ago

How many grams of sucrose (C12H22O11) would be in 8.7 L of a 6.81M solution of sucrose

Chemistry

1 answer:

VladimirAG [237]3 years ago

7 0

<h3><u>Answer;</u></h3>

= 20,280.25 g or 20.28 Kg

<h3><u>Explanation;</u></h3>

Number of moles, n = Molarity × Volume

Molarity = 6.81 M

Volume = 8.7 L

Number of moles = 6.81 × 8.7

= 59.247 moles

Mass = Moles × Molar mass

= 59.247 × 34.2 g/mol

= 20,280.25 or

= 20.28 Kg

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How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg

Afina-wow [57]

Answer : The mass of $PbSO_4$ needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of $PbCrO_4$ .

$\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole$

Now we have to calculate the moles of $PbSO_4$ .

The balanced chemical reaction will be,

$PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4$ produced from 1 mole of $PbSO_4$

So, 0.005 mole of $PbCrO_4$ produced from 0.005 mole of $PbSO_4$

Now we have to calculate the mass of $PbSO_4$

$\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4$

$\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g$

Therefore, the mass of $PbSO_4$ needed are, 1.515 grams.

6 0

3 years ago

1-magma wich flows through channles to the surface will produce a???

AlladinOne [14]

Answer:

thats one hint. but for 1 it will produce lava

Explanation:

number 2

some of the Earth's grandest mountains are composite volcanoes--sometimes called stratovolcanoes. They are typically steep-sided, symmetrical cones of large dimension built of alternating layers of lava flows, volcanic ash, cinders, blocks, and bombs and may rise as much as 8,000 feet above their bases

6 0

4 years ago

7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s

Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both $NH_{3}$ and $CO_{2}$ are gaseous. Hence equilibrium constant depends upon partial pressures of $NH_{3}$ and $CO_{2}$ .

Initially no $NH_{3}$ and $CO_{2}$ were present.

Hence mole fraction of $NH_{3}$ and $CO_{2}$ at equilibrium can be calculated from coefficient of $NH_{3}$ and $CO_{2}$ in balanced equation.

Mole fraction of $NH_{3}$ = (number of moles of $NH_{3}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{2moles}{(2+1)moles}=\frac{2}{3}$

Mole fraction of $CO_{2}$ = (number of moles of $CO_{2}$ )/(total number of moles of $NH_{3}$ and $CO_{2}$ ) = $\frac{1moles}{(2+1)moles}=\frac{1}{3}$