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BigorU [14]
3 years ago
8

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}

Let's assume both CO_{2} and NH_{3} behaves ideally.

Therefore partial pressure of NH_{3}, P_{NH_{3}}= x_{NH_{3}}.P_{total} and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm

P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm

So, K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888

4 0
3 years ago
6. If 981 mL of a gas is heated from 335 K at constant pressure and the volume
storchak [24]

Answer:

1.5L

Explanation:

The new volume can be found by using the formula for Boyle's law which is

p1v1=p2v2

Since we are finding the new volume

v2= p1v1/p2

From the question we have

v2= 5x3/10=15/10= 3/5

We have the final answer as

1.5 L

3 0
3 years ago
What are products of a reaction between KOH(aq) and HCI(aq)
nata0808 [166]

Answer:

salt and water only

Explanation:

neutralization reaction

8 0
3 years ago
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