When you rub an inflated balloon on your head and it makes your hair stand up, the force that makes the hair stand up is known as static electricity.
When the balloon is rubbed on the head, electrons from the hair atoms move into the balloon, thus making the balloon to be negatively charged and the hair positively charged due to loss of electrons.
Unlike charges attracts. Thus, when you try to pull the balloon away slowly, the positively charged hair and the negatively charged balloon will attract each other and this is usually what makes the hair stand up.
More on static electricity can be found here: brainly.com/question/24160155
Answer:
B
Explanation:
the pH scale is based on the function
1x10-14
So since pH is neutral the concentration of OH and H must be equal hence the only option in which the concetration would [OH]{H}= 1x10-14 would be when both are equal so it cannot be a or d and when they are expressed in concetration so the only option is B.
The reaction to form NH3 is : N2 + 3H2-> 2NH3 12,33g NH3 is 12,33/17,03=0,3 =0,724 moles of NH3 moles NH3. So you need 1,5*0,724 = 1,086 moles H2 1,086*2,016 = 2,189 g of H2 is needed ro form 12,33 g NH3
When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:
(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃
Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
The change in pH is calculated by:
pOH = Protein kinase B + log [NH4+]/ [NH3]
Protein kinase B of ammonia = 4.74
initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74
pH = 14 - 4.74=9.26
moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300
NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.000300=0.00970
moles NH4+ = 0.0100 + 0.000300=0.0103
pOH = 4.74 + log 0.0103/ 0.00970= 4.77
oH = 14 - 4.77 = 9.23
the change is = 9.26 - 9.23 =0.03
