The heat that is needed to raise the temperature of 78.4 g of aluminium from 19.4 °c to 98.6°c is 5600.77 j
<u><em>calculation</em></u>
Heat(Q) = mass(M) x specific heat capacity (C) x change in temperature(ΔT)
where;
Q=?
M = 78. 4 g
C=0.902 j/g/c
ΔT=98.6°c -19.4°c =79.2°c
Q is therefore = 78.4 g x 0.902 j/g/c x 79.2°c =5600.77 j
Answer:
Requirements for a correctly written chemical equation are reactants and products, their formula and valency
Explanation:
Formula of the given compound are -
1 - Potassium Hydroxide - 
2 - Calcium Nitrate - 
The requirements for a correctly written chemical equation are -
- Identifying reactants and products
- Formula of reactants and products
- Valency of elements
Example of word equation, formula equation, and chemical equation is as follows -
Aluminium + iron9(III)oxide ⇒ aluminium oxide + iron (word equation)
+ 
⇒ 
+ 
(formula equation)
+ ⇒ + 
(chemical equation)
Answer:
0.55 mol Au₂S₃
Explanation:
Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.
1. Gather all the information in one place:
M_r: 34.08
Au₂S₃ + … ⟶ 3H₂S + …
m/g: 56
2. Calculate the moles of H₂S
Moles of H₂S = 56 g H₂S × (34.08 g H₂S/1 mol H₂S)
= 1.64 mol H₂S
3. Calculate the moles of Au₂S₃
The molar ratio is 1 mol Au₂S₃/3 mol H₂S.
Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)
= 0.55 mol Au₂S₃
Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² / 0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.
Answer:
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