Answer:
B. Poor conductor.
Explanation:
It cannot be A, as only 1 metal is not solid at room temp.
It cannot be C, as most metals are ductile.
It cannot be D, as most metals are malleable.
This leaves B, which is not true about metals, as a lot are very good conductors.
Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!
We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.
Answer:
5.81 moles
Explanation:
To find the number of moles (n) in 3.5 x 10²⁴ molecules of methane gas, we divide the number of molecules by Avagadro's number (nA). That is,
n = number of molecules ÷ 6.02 × 10²³
According to this question, 3.5 x 10^24 molecules of methane gas was given, hence,
n = 3.5 × 10²⁴ ÷ 6.02 × 10²³
n = 3.5/6.02 × 10(24 - 23)
n = 0.5814 × 10¹
n = 5.81 moles
