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3 years ago

30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?

1 answer:

5 0

Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

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Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq

Rasek [7]

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s) + 2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

7 0

3 years ago

Read 2 more answers

Cuántos moles hay en 17.5 gramod de ZnCl2

Aliun [14]

Explanation:

Zn=65

Cl2= 35+35=70

65+70=135g

1 mole ZnCl2 = 135g

x mole = 17.5g

17.5g × 1 mole/ 135g= 0.129 moles en 17.5g de ZnCl2

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3 years ago

Mercury (Hg) poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes l

alisha [4.7K]

Answer:

6 x 10⁵ kg Hg

Explanation:

The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.

The volume of the lake is calculated in cubic feet:

V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³

Cubic feet are then converted to mL (1cm³=1mL)

(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL

The mass of mercury is then found:

m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg

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A compound is 2% H, 32.7% S, and 65.3% O by mass. What is the subscript on the O in the empirical formula for this compound?

IceJOKER [234]

Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped!

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