Good electrical conductivity and electronegativities less than 1.7 are the properties and characteristic of Group 2 elements at STP.
<h3>What are the properties of group 2 elements?</h3>
Group 2 elements are metals so they are good conductors of heat and electricity. It has electronegativity values less than 1.7 and very reactive. They form 2+ charge in cationic form and also formed ionic bonds with other negatively charged elements.
So we can conclude that good electrical conductivity and electronegativities less than 1.7 are the properties and characteristic of Group 2 elements at STP.
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Answer:
C
Explanation:
A diatomic element in that list is Bromine
Answer:
a. The conjugate base of an acidic buffer will accept hydrogen protons when a strong acid is added to the solution.
b. An acidic buffer solution is a mixture of a weak acid and its conjugate base.
e. The weak acid of an acidic buffer will donate hydrogen protons when a strong base is added to the solution.
Explanation:
<em>Which of the statements correctly describe the properties of a buffer?</em>
a. The conjugate base of an acidic buffer will accept hydrogen protons when a strong acid is added to the solution. TRUE. The conjugate base neutralizes the excess of hydrogen protons.
b. An acidic buffer solution is a mixture of a weak acid and its conjugate base. TRUE.
c. An acidic buffer solution is a mixture of a weak base and its conjugate acid. FALSE. This is a basic buffer solution.
d. The weak acid of an acidic buffer will accept hydrogen protons when a strong base is added to the solution. FALSE. The weak acid will react with the hydroxyl ions from the added base.
e. The weak acid of an acidic buffer will donate hydrogen protons when a strong base is added to the solution. TRUE. These hydrogen protons will form water.
f. The conjugate base of an acidic buffer will donate hydrogen protons when a strong acid is added to the solution. FALSE. It will accept hydrogen protons.
Answer: Boiling chips provide surfaces on which bubbles can form as the liquid boils.
Hope this helps!
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :
when Q = M*C*ΔT
Q is the heat in J
and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g
C is the specific heat J/g.°C
ΔT is the change in temperature
Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]
= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]
∴Q = 7444.8 J
and when we know that the heat of fusion for water = 334J/g
and heat of vaporization for water = 2260J/g
∴Q for the two phases changes = M * (2260+334)
= 18 * (2260+334)
= 46692 J
∴ Q total = 7444.8 + 46692 = 54136.8 J
