When 189.6 g of ethylene (C2H4) burns in oxygen to give carbon dioxide and water, how many grams of CO2 are formed?

820 g Li2SO4 is dissolved into 2500 mL of solution. What is the molar concentration?

Andreas93 [3]

Molarity is defined as the moles of solute per liter of solution. $M=\frac{n}{V}$ . Where M is molarity, n is the number of moles and V is the volume. First we must find the molar mass of $Li_2SO_4$ which is 109.98 g/mol

$Li_2SO_4= 2 \times Li_{Ar} + S_{Ar} + 4 \times O_{Ar}\\= (2 \times 6.941 + 32.1 + 4 \times 16.0) = 109.98 g/mol$

$820\ g\ Li_2SO_4 \times \frac{mol}{109.98\ g}= 7.46\ mol\ Li_2SO4$

Then we find the molarity using above equation

$M= \frac{n}{V} = \frac{7.46 mol}{2500ml} \times\frac{1000ml}{L} = 2.98\ M$

6 0

3 years ago

How many significant figures are in 5.40

kotykmax [81]

There are 3 significant figures in this value, all values before and after the decimal point are significant. As there is a decimal point, the zeros trailing are also significant.

4 0

3 years ago

What is the VSEPR model of CB4I?

sesenic [268]

Answer:

Molecular geometry Vsepr

According to VSEPR, the valence electron pairs surrounding an atom mutually repel each other; they adopt an arrangement that minimizes this repulsion, thus determining the molecular geometry. This means that the bonding (and non-bonding) electrons will repel each other as far away as geometrically possible.

Explanation:

3 0

3 years ago

How did you use QPOE?

Jlenok [28]

QPOE Files

The x-ray data are stored in QPOE files (Quick Position-Ordered Events, *.qp) rather than image arrays. These are lists of photons identified by several quantities, including the position on the detector, pulse height, and arrival time. Note that, unlike IRAF images, QPOE files have no associated header file, and are always stored in the current directory, unless explicitly specified otherwise. Non-PROS IRAF tasks can also access QPOE data files in place of image arrays.

8 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago