Answer:
1. atomic #
2.The sugar-water is a homogeneous mixture
Explanation:
1. However, if it has positive ion, then this electron number will go down (ie +2 charge means two electrons have been lost, so the electron/atomic number will go down by two) and vice versa.
2.Sugar dissolves and is spread throughout the glass of water. The sand sinks to the bottom. The sugar-water is a homogenous mixture while the sand-water is a heterogeneous mixture. Both are mixtures, but only the sugar-water can also be called a solution.
Answer:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Explanation:
We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.
HNO₃ is a strong acid and will have the largest hydronium concentration.
HCN Ka = 6.2 x 10⁻¹⁰
HNO₂ Ka = 4.0 x 10⁻⁴
HClO Ka = 3.0 x 10⁻⁸
The ranking from smallest to largest hydronium concentration will then be:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Answer is: <span>yield of a reaction is 56,4%.
</span>Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
m(PCl₃) = 200 g.
m(HCl) = 91,0 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 200 g ÷ 137,33 g/mol.
n(PCl₃) = 1,46 mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 91 g ÷ 36,45 g/mol.
n(HCl) = 2,47 mol.
From reaction: n(PCl₃) : n(HCl) = 1 : 3.
n(HCl) = 1,46 mol · 3 = 4,38 mol.
Yield of reaction: 2,47 mol ÷ 4,38 mol · 100% = 56,4%.
The value of "d" is 80°
Explanation:
Cyclic quadrilaterals are the special group of quadrilaterals with all its base lying on the circumference of the circle. In other words, a quadrilateral inscribed in a circle is called a cyclic quadrilateral.
Cyclic quadrilateral are characterised by some special features such as
- Sum of opposite angles of a cyclic quadrilateral is always a supplementary angle.
- If one of the sides of a cyclic quadrilateral is produced, then the exterior angle so formed is always double of the corresponding interior angle.
Using the property 1
We find that since the quadrilateral is cyclic, opposite pairs must be supplementary
100°
+∠D must be equal to 180°
D=180°
-100°
=80°
