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trasher [3.6K]
2 years ago
14

How many moles of lithium are present in a sample that contains 2.45x10^87 formula units of Li2SO4?​

Chemistry
1 answer:
ddd [48]2 years ago
3 0

The answer is: 8.14·10⁶³ moles of lithium are present.

N(Li₂SO₄) = 2.45·10⁸⁷; number of formula units of lithium sulfate.

n(Li₂SO₄) = N(Li₂SO₄) ÷ Na.

n(Li₂SO₄) = 2.45·10⁸⁷ ÷ 6.022·10²³ 1/mol.

n(Li₂SO₄) = 4.07·10⁶³ mol; amount of lithium sulfate

In one molecule of lithium sulfate, there are two atoms of lithium.

n(Li₂SO₄) : n(Li) = 1 : 2.

n(Li) = 2 · 4.07·10⁶³ mol.

n(Li) = 8.14·10⁶³ mol; amount of lithium atoms.

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Brian's aunt has cats. When Brian recently visited her, he started sneezing badly and believes that it was because
kifflom [539]

Answer:

By visiting other households with cats.

Explanation:

This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.

4 0
2 years ago
The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl),
choli [55]

Answer:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

B. 100.00° C

C. 101.02° C

D. 100.51° C

Which of the following solutions will have the lowest freezing point?

A. 1.0 mol/kg sucrose (C12H22O11)

B. 1.0 mol/kg lithium chloride (LiCl)

C. 1.0 mol/kg sodium phosphide (Na3P)

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

B. magnesium sulfate (MgSO4)

C. potassium bromide (KBr)

D. All will be equally effective.

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

A. 98.7 kPa

B. 96.3 kPa

C. 101.3 kPa

D. 100.2 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

A. 101.2°C

B. 105.9°C

C. 102.7°C

D. 108.1°C

Explanation:

The following are the answers to the different questions: 

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

Which of the following solutions will have the lowest freezing point?

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

B. 96.3 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

D. 108.1°C

5 0
2 years ago
At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?
Gnom [1K]
The temperature that  would  the volume of a gas  be 0.550l  if  it  had a volume of 0.432 L  at  -20.0  c is calculated  using the Charles law formula

that is   v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l

by  making T1  the subject of the formula  T1= V1T2/V2


T1=  (0.55lL x253)/  0.432 l = 322.11 K  or  322.11-273 = 49.11 C
8 0
2 years ago
Read 2 more answers
50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.
photoshop1234 [79]

Limiting reactant : O₂

Mass of  N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

\tt =\dfrac{50}{44}=1.136

mol O₂ :

\tt =\dfrac{50}{32}=1.5625

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136    1.5625

1.0416  1.5625    1.0416

0.0944    0          1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

\tt =mol\times MW=1.0416\times 92=95.83~g

7 0
2 years ago
How many grams of (nh4)3po4 are needed to make 0.250 l of 0.150 m (nh4)3po4? hints how many grams of (nh4)3po4 are needed to mak
NeX [460]
<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. We calculate the mass of the solute by first determining the number of moles needed. And by using the molar mass, we can convert it to units of mass.

Moles </span>(nh4)3po4 = 0.250 L (0.150 M) = 0.0375 moles (nh4)3po4
Mass = 0.0375 mol (nh4)3po4 (149.0867 g / mol) = 5.59 g (nh4)3po4
5 0
2 years ago
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