The percentage of sugar in the solution is 1%
To solve this question, we need to find the percentage of 6 out of 600.
Data given;
- mass of sugar = 6g
- mass of solution = 600g
<h3 /><h3>What is Percentage</h3>
This is the rate or proportion of a substance in the total value and multiplied by 100.
The percentage of sugar in the solution is calculated by
The percentage of sugar in the solution is 1%
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Answer:
5.1 Crystal structure and habit.
5.2 Hardness.
5.3 Lustre and diaphaneity.
5.4 Colour and streak.
5.5 Cleavage, parting, fracture, and tenacity.
5.6 Specific gravity.
5.7 Other properties.
Answer:
d it increase the number of collisions
Explanation:
cos collision is a type reaction to other particles
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Answer:
(a) The rate of formation of K2O is 0.12 M/s.
The rate of formation of N2 is also 0.12 M/s
(b) The rate of decomposition of KNO3 is 0.24 M/s
Explanation:
(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.
Rate of formation of O2 is 0.3 M/s
Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s
Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.
Rate of formation of N2 = (2×0.3/5) = 0.12 M/s
(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.
Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s
Answer:
-285.4 J/K
Explanation:
Let's consider the following balanced equation.
HCl(g) + NH₃(g) ⇒ NH₄Cl(s)
We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.
ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))
ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol
ΔS°r = -285.4 J/K