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1 year ago

How many chloride ions are in a 220 grams of calcium chloride?

2 answers:

6 0

The Ions present in CaCl₂ are,

Ca²⁺ Cl⁻ Cl⁻

Means 1 formula unit contains 1 Ca²⁺ ion and 2 Cl⁻ ions.

Also, 1 mole of CaCl₂ contains 6.022 × 10²³ formula units.

So, 1 mole formula units of CaCl₂ contain,

2 × 6.022 × 10²³ = 1.20 × 10²⁴ Cl⁻ Ions

Now, Calculating number of moles contained by 220 g of CaCl₂,
As,
110.98 g of CaCl₂ = 1 mole
Then,
220 g of CaCl₂ = X moles
Solving for X,
X = (220 g × 1 mol) ÷ 110.98 g

X = 1.98 moles
As,
1 mole contained = 1.20 × 10²⁴ Cl⁻ Ions
Then,
1.98 mole will contain = X Cl⁻ Ions
Solving for X,
X = (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol

X = 2.38 × 10²⁴ Cl⁻ Ions

AURORKA [14]1 year ago

3 0

Answer:

10.24 x 10^24 ions. i just took the test. have a great day

Explanation:

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A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

5 0

1 year ago

(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path

never [62]

Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.

4 0

2 years ago

Read 2 more answers

The Lewis structure for CO2 has a central The Lewis structure for C O 2 has a central blank atom attached to blank atoms.

kirza4 [7]

Answer:

See the explanation

Explanation:

1) The Lewis structure for $CO_2$ has a central Carbon atom attached to Oxygen atoms.

In the $CO_2$ we will have a structure: O=C=O the central atom "carbon" we will have 2 sigma bonds and 2 pi bonds, therefore, we have an Sp hybridization. For O we have 1 pi and 1 sigma bond, therefore, we have an Sp2 hybridization.

2) These atoms are held together by double bonds.

Again in the structure of $CO_2$ : O=C=O we only have double bonds.

3. Carbon dioxide has a Carbon dioxide has a Linear electron geometry.

Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.

4. The carbon atom is Sp hybridized.

We will have for carbon 2 pi bonds, so we will have an Sp hybridization.

5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.

(See figures)

Figure 1: Carbon hybridization

Figure 2: Oxygen hybridization

6 0

2 years ago

At a given set of conditions 241.8 kJ of heat is released when one mole of H2O forms from its elements. Under the same condition

yan [13]

Answer:

44Kj

Explanation:

These are the equations for the reaction described in the question,

Vaporization which can be defined as transition of substance from liquid phase to vapor

H2(g)+ 1/2 O2(g) ------>H2O(g). Δ H

-241.8kj -------eqn(1)

H2(g)+ 1/2 O2(g) ------>H2O(l).

Δ H =285.8kj ---------eqn(2)

But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as

H2O(l) ------>>H2O(g)---------------eqn(3)

But the equation from eqn(2) the eqn does go with vaporization so we can re- write as

H2O ------> H2(g)+ 1/2 O2(g)

Δ H= 285.8kj ---------------eqn(4)

To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)

Δ H=285.8kj +(-241.8kj)= 44kj

5 0

1 year ago

An example of a physical property of an element is the element’s ability to(1) react with an acid(2) react with oxygen(3) form a

ivanzaharov [21]

4 sounds correct......

4 0

2 years ago