Answer:
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Answer:
Moisture content in wet basis = 47.4 %
Explanation:
Moisture content expresses the amount of water present in a moist sample.Dry basis and wet basis are widely used to express moisture content.
The next equation express the moisture content in wet basis:
where, : moisture content in wet basis and
: moisture content in dry basis
We now calculate the moisture content in wet basis:
= 0.474 = 47.4 % wet basis
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0.085 moles of Al are required to form 23.6 g of AlBr₃.
Let's consider the following balanced equation for the synthesis reaction of AlBr₃.
2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)
First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).
The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885 moles of AlBr₃ are:
0.085 moles of Al are required to form 23.6 g of AlBr₃.
You can learn more about stoichiometry here: brainly.com/question/22288091
Answer:
0.088 mole of Al.
Explanation:
First, we shall determine the number of mole in 23.6 g of AlBr₃.
This is illustrated below:
Mass of AlBr₃ = 23.6 g
Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol
Mole of AlBr₃ =.?
Mole = mass/Molar mass
Mole of AlBr₃ = 23.6 / 267
Mole of AlBr₃ = 0.088 mol
Next, we shall writing the balanced equation for the reaction.
This is given below:
2Al(s) + 3Br₂(l) → 2AlBr₃(s)
From the balanced equation above,
2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.
Finally, we shall determine the number of mole of Al needed for the reaction as follow:
From the balanced equation above,
2 moles of Al reacted to 2 moles AlBr₃.
Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.
B.) Positively charged
C.) Strongly ionic
D.) Negatively charged
Answer with double replacement
Answer:
Barium chloride + Sodium phosphate → barium phosphate + sodium chloride
Explanation:
Double replacement:
It is the reaction in which two compound exchange their ions and form new compounds.
AB + CD → AC +BD
Chemical equation:
BaCl₂ + Na₃PO₄ → Ba₃(PO₄)₂ + NaCl
Balanced chemical equation:
3BaCl₂ + 2Na₃PO₄ → Ba₃(PO₄)₂ + 6NaCl
The cation and anion of both reactants are exchanged with each other.
Ba²⁺ react with PO₄³⁻ and form Ba₃(PO₄)₂ while Cl⁻ react with Na⁺ and form sodium chloride.
Molecular equation:
Barium chloride + Sodium phosphate → barium phosphate + sodium chloride
A double replacement reaction between barium chloride and sodium phosphate results in the formation of barium phosphate and sodium chloride.
The given chemical equation represents a double replacement reaction. The chemical reactants are barium chloride (BaCl2) and sodium phosphate (Na3PO4). In a double replacement reaction, the cations and anions of the two reactants switch places to form two new compounds.
So here is how the reaction would proceed: BaCl2 + Na3PO4 -> Ba3(PO4)2 + NaCl.
This translates to: Barium chloride reacts with sodium phosphate to form barium phosphate and sodium chloride.
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Answer:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
α =2
β = 19
γ = 12
δ = 14
53.2moles of O₂
Explanation:
Proper equation of the reaction:
αC₆H₁₄ + βO₂ → γCO₂ + δH₂O
This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:
CₓHₙ + (x + )O₂ → xCO₂ + H₂O
From the given combustion equation, x = 6 and n = 14
Therefore:
β = x + = 6 + = 6 + 3.5 = 9
γ = 6
δ = = = 7
The complete reaction equation is therefore given as:
C₆H₁₄ + 9O₂ → 6CO₂ + 7H₂O
To express as whole number integers, we multiply the coefficients through by 2:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
Problem 2
From the reaction:
2 moles of hexane are required to completely react with 19 moles of O₂
∴ 5.6 moles of hexane would react with k moles of O₂
This gives: 5.6 x 19 = 2k
k =
k = 53.2moles of O₂
What is the molarity?
Answer: The concentration of is 0.0122 M.
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
Hence, the concentration of is 0.0122 M.