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Alenkasestr [34]
3 years ago
7

Help me with this please i need chem help lol

Chemistry
1 answer:
melisa1 [442]3 years ago
7 0

Answer:

\boxed{\text{1. 0.47 mol; 2. 0.046 mol; 3 (a) 11.6 mol, 15.4 mol, (b) 2.31 mol, 3.08 mol, 0.770 mol}}

Explanation:

1. Moles of Na

(a) Balanced equation

2Na + 2H₂O ⟶ 2NaOH + H₂

(b) Calculation

You want to convert moles of H₂ to moles of Na

The molar ratio is 2 mol Na:1 mol H₂

Moles of Na  = 4.0 mol H₂ × (2 mol Na/1 mol H₂) = 8.0 mol Na

You need \boxed{ \text{8.0 mol of Na}} to form 4.0 mol of H₂.

2. Moles of LiCl

(a) Balanced equation

2LiBr + Cl₂⟶ 2LiCl + Br₂

(b) Calculation

You want to convert moles of LiBr to moles of LiCl

The molar ratio is 2 mol LiBr:2 mol LiCl

Moles of LiCl  = 0.046 mol LiBr × (2 mol LiCl/2 mol LiBr) = 0.046 mol LiCl

The reaction will produce \boxed{ \text{0.046 mol of LiCl}}.

3. Combustion of propane

C₃H₈ +5O₂ ⟶ 3CO₂ +4H₂O

(a) Moles of CO₂ and H₂O

Moles of CO₂ = 0.647 mol O₂ × (3 mol CO₂/1 mol C₃H₈) = 11.6 mol CO₂

Moles of H₂O = 3.85 mol O₂ × (4 mol CO₂/1 mol C₃H₈) = 15.4 mol H₂O

The reaction produces \boxed{ \text{11.6 mol of CO}_{2}} and \boxed{ \text{15.4 mol of H}_{2}\text{O}}.

(b) Moles from O₂

Moles of CO₂ = 3.85 mol O₂ × (3 mol CO₂/5 mol O₂) = 2.31 mol CO₂

Moles of H₂O = 3.85 mol O₂ × (4 mol CO₂/5 mol O₂) = 3.08 mol H₂O

Moles of C₃H₈ = 3.85 mol O₂ × (1 mol C₃H₈/5 mol O₂) = 0.770 mol C₃H₈

The reaction produces \boxed{ \text{2.31 mol}} of CO₂, \boxed{ \text{3.08 mol}} of H₂O, and consumes \boxed{ \text{0.770 mol}} of C₃H₈.

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Fudgin [204]

Answer:

  • <u><em>Magnesium and fluorine.</em></u>

Explanation:

<em>Ionic compounds</em> are formed by the electrostatic attraction of cations and anions.

Cations, positive ions, are formed when atoms lose electrons, and anions, negative ions, are formed when atoms gain electrons.

When two different atoms have similar atraction for electrons (electronegativity) they will not donate to nor catch electrons from each other, so cations and anions will not be formed. Instead, the atoms would prefer to share electrons forming covalent bonds to complete their outermost shell (octet rule).

Then, in order to form ionic compounds the electronegativities have to substantially different. This situation does not happen between two nonmetal elements, which nitrogen and sulfur are. Then, you can predict safely that nitrogen and sulfur will not form an ionic compound.

Ionic compounds, then require the electronegativity difference that exist between some metals and nonmetals. Being magnesium an alkaline earth metal, its electronegativity is very low. On the other hand, fluorine the first element of the group 17, has the highest electronegativity of all the elements.Thus magnesium and fluorine will have enough electronegativity difference to justify the exchange of electrons, forming ions and, consequently, ionic compounds.

5 0
3 years ago
If you have 5 moles of carbon dioxide at 30degrees C and under 600mmHg of pressure, what will the volume be
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Answer:

157.64 L

Explanation:

We'll begin by converting 30 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 30 °C

T(K) = 30 °C + 273

T (K) = 303 K

Next, we shall convert 600 mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

600 mmHg = 600 mmHg × 1 atm / 760 mmHg

600 mmHg = 0.789 atm

Finally, we shall determine the volume of the gas. This can be obtained as follow:

Number of mole (n) = 5 moles

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Pressure (P) = 0.789 atm

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0.789 × V = 5 × 0.0821 × 303

0.789 × V = 124.3815

Divide both side by 0.789

V = 124.3815 / 0.789

V = 157.64 L

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2 years ago
In the lab you measure a clean dry crucible and cover to be 24.36 grams. You obtain a 2cm piece of pure magnesium metal. After s
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Answer:

1) 0.3g Mg

2)0.5g MgO

3)0.2g O

4)0.01mol Mg & 0.01mol O

5)0.01mol MgO

6) Empirical formula MgO

Explanation:

The mass og Mg is obtained by substracting 24.36g from 24.66g:

24.66 - 24.36 = 0.3g Mg

The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.

We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:

0.3g Mg*\frac{16gO}{24.3gMg}= 0.2g O

Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO

We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:

0.2g O*\frac{1 mol O}{16g O}= 0.01mol O

0.3g Mg*\frac{1mol Mg}{24.3g Mg}= 0.01mol Mg

The moles of MgO can be obtained from:

0.5g MgO*\frac{1mol MgO}{40.3g MgO}= 0.01mol MgO

To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.

The result for both number of  Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the  formula unit of the compound.

The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.

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Answer:

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