Answer:
x = 1, -7.5
Explanation:
2x² + 13x = 15
Divide the left side of the equation by 2
2(x² + 6.5x) = 15
Divide 6.5 by 2 and square the quotient
6.5/2 = 3.25
3.25² = 10.5625
Add 10.5625 to the left side
2(x² + 6.5x + 10.5625) = 15
Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.
10.5625 • 2 = 21.125
2(x² + 6.5x + 10.5625) = 36.125
To factor (x² + 6.5x + 10.5625), add b/2 to x
b/2 = 6.5/2 = 3.25
2(x + 3.25)² = 36.125
Divide by 2
(x + 3.25)² = 18.0625
Square root.
(x + 3.25) = √18.0625
x + 3.25 = ±4.25
Subtract 3.25.
x = 4.25 - 3.25 = 1
x = -4.25 - 3.25 = -7.5
x = 1, -7.5
Explanation:
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Explanation:
The given data is as follows.
Concentration of standard NaOH solution = 0.1922 M
Let the original acid solution concentration be x.


= 0.1 x M
= 10.00 mL (given)
The reaction equation is as follows.

Concentration × Volume of
= Concentration × Volume of NaOH

x = 1.314 M
Therefore, we can conclude that the concentration of the original sulfuric acid solution is 1.314 M.
Answer: A. The compound CuCl is 500 times less soluble in sea water than it is in pure water.
pure water.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in a equilibrium reaction, the equilibrium will shift in a direction so as to minimize the effect.
Thus when a common ion is introduced to an equilibrium reaction, the equilibrium will shift in a direction where the concentration of common ion is decreasing.


When common ion such as
from NaCl in sea water is introduced to an equilibrium reaction, the equilibrium will shift in a direction where the concentration of common ion is decreasing i.e. in the left side and thus solubility of CuCl further decreases.
On the periodic table, hydrogen and helium are the only two elements in the first row, or period, which reflects that they only have electrons in their first shell. Hydrogen and helium are the only two elements that have electrons exclusively in the 1 s 1s 1s orbital in their neutral, non-charged, state.