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Ede4ka [16]
2 years ago
10

What is the mass of 2.50L of SO2 gas at stp

Chemistry
2 answers:
maxonik [38]2 years ago
5 0
STP is abbreviation for Standard Temperature and Pressure at which the temperature is 273 K and pressure is 1 atm 
- At these conditions the molar volume is equal to 22.4 L
so 1 mole of SO₂ volume = 22.4 L 
     ? mole of SO₂ volume = 2.5 L
number of moles = 2.5 / 22.4 = 0.1116 mol
mass of SO₂ = 0.1116 * 64.063 = 7.15 g
stira [4]2 years ago
4 0

Answer: 4.84 grams

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass, occupies 22.4 L of volume at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given volume}}{\text{Molar volume}}=\frac{2.50}{22.4}=0.11

1 mole of SO_2 weighs = 44 g

0.11 moles of SO_2 weigh = \frac{44}1}\times 0.11=4.84g

Thus mass of 2.5 L of SO_2 weigh 4.84 grams.

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If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

8 0
3 years ago
A student conducts an experiment by placing a wooden stick in a cup of sugar solution. Over the next few weeks, sugar crystals f
Ipatiy [6.2K]

Answer: In order to be alive there must be cells present and since sugar does not contain cells it is not alive.

Explanation:

8 0
2 years ago
In what two places minerals are form?
larisa [96]
In earths surface or the bottom of the Ocean
3 0
3 years ago
A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the
BARSIC [14]

Answer:

Rb+

Explanation:

Since they are telling us that the equivalence point was reached after 17.0 mL of   2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.

Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and  we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n,  of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.

Thus our calculations are:

V = 17.0 mL x 1 L / 1000 mL = 0.017 L

2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol

0.0425 mol = 4.36 g/ MW XOH

MW of XOH = (atomic weight of X + 16 + 1)

so solving the above equation we get:

0.0425 = 4.36 / (X + 17 )

0.7225 +0.0425X = 4.36

0.0425X = 4.36 -0.7225 = 3.6375

X = 3.6375/0.0425 = 85.59

The unknown alkali is Rb which has an atomic weight of 85.47 g/mol

6 0
3 years ago
The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.
Naddika [18.5K]

Answer:

12.6.

Explanation:

  • We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

  • It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

4 0
3 years ago
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