A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is
.
Explanation:
The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.
Now, in terms of chemical formulae this reaction equation will be as follows.

Here, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
To balance this equation, multiply HCl by 4 on reactant side and multiply
by 2 on product side. Therefore, the equation can be rewritten as follows.

Hence, number of atoms on reactant side are as follows.
Number of atoms on product side are as follows.
Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.
Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is
.
(also called Observational Error) is the difference between a measured quantity and its true value. It includes random error
Answer:
solubility of X in water at 17.0
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0
can be calculated using the information given.
Let's assume solubility of X in water at 17.0
is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y =
= 0.11
Hence solubility of X in water at 17.0
is 0.11 g/mL.
Answer:
a) [H3O+] = 1.00 E-10 M ⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M ⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M ⇒ [OH-] = 1.0 E-9 M
Explanation:
- 14 = pH + pOH
- pH = - Log [H3O+]
a) [H3O+] = 1.00 E-10 M
⇒ pH = - Log(1.00 E-10) = 10
⇒ pOH = 14 - 10 = 4
⇒ 4 = - Log[OH-]
⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M
⇒ pH = 4
⇒ pOH = 10
⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M
⇒ pH = 5
⇒ pOH = 9
⇒ [OH-] = 1.0 E-9 M