Answer:
A catalyst is a chemical substance that alters the rate of chemical reaction not consumed by the reaction. Hence, a catalyst can be recovered chen unchanged at the ends of chemical reaction. Catalyst can be divided into two typ the basis whether it speeds up or slowdowns the rate of chemical reaction. The positive catalyst and negative catalyst.
Answer:
ΔHrxn = - 1534.3 J
Explanation:
Given the assumptions and the formula for the change in enthalpy:
ΔHrxn = m x C x ΔT, where
m is the mass of solution given 135.4 g
C is the heat capacity 4.2 J/g .K and,
ΔT is the change in temperature
we have ,
T₁ = ( 18.1 + 273) K = 291.1 K
T₂ = ( 15.4 +273) K = 288.4 K
ΔHrxn = 135.3 g x 4.2 J/gK x ( 288.4 -291.1 ) K = - 1534.3 J
After verifying our result has the correct unit, the answer is -1534.3 Joules, and the negative sign tells us it is an endothermic reaction decreasing the final temperature.
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
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Explanation:
Scientifically speaking, metals are naturally occurring chemical elements that are typically hard, lustrous, and good conductors of both heat and electricity. Examples include iron, gold, silver, copper, zinc, nickel, etc., but also elements we don't normally think of as metals.
When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.
<h3>What is a precipitate?</h3>
The precipitate is the solid concentration of a substance that is collected over a solution.
First, we determine the concentration of magnesium and hydroxide
(Mg2+) = 7.00 × 10⁻⁴
(OH−) = 5.00 × 10⁻⁵
Now, we calculate the solubility quotient
Qc = (Mg2+) (OH−) ²
Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²
Qc = 1.75 x 10⁻¹²
The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.
Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.
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