Answer:
Explanation:
Pv = n RT
P =7.65, T= 310K, for glucose n = moles= weight/molar mass= weight/180, R is gas constant
7.65* V =W/180 *0.0821 * 310
w/v= 7.65*180/0.0821*310 = 1377/25.45 = 54.10 =5.4%
Answer:
9.4L
Explanation:
Given parameters:
Mass of Cl₂ gas = 30g
Unknown:
Volume of gas = ?
Solution:
To solve this problem, let us assume that this reaction takes place under standard temperature and pressure conditions.
Therefore;
Volume of gas = number of moles x 22.4
To find the number of moles;
Number of moles =
molar mass of Cl₂ = 2(35.5) = 71g/mol
Number of moles = = 0.42mole
So;
Volume of gas = 0.42 x 22.4 = 9.4L
Answer is: volume is 25.08 liters.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂(g).
m(Na) = 51.5 g; mass of sodium.
n(Na) = m(Na) ÷ M(Na).
n(Na) = 51.5 g ÷ 23 g/mol.
n(Na) = 2.24 mol; amount of sodium.
From chemical reaction: n(Na) : n(H₂) = 2 : 1.
n(H₂) = 2.24 mol ÷ 2.
m(H₂) = 1.12 mol; amount of hydrogen gas.
V(H₂) = n(H₂) · Vm.
V(H₂) = 1.12 mol · 22.4 L/mol.
V(H₂) = 25.08 L; volume of hydrogen gas.
Ionization enthalpy, IE, is also called ionization potential is the ability to remove the electron from the neutral gaseous atom. There is a trend observed in the periodic table for the IE value. As we go from left to right in a period, IE vale increases. While moving from top to bottom in a group, IE value decreases.
- The phenomenon of unexpected drop in IE1 values between Groups 2 and 13, in period 2 and period 4 is due to the introduction of d-orbitals in the case of period 4 elements.
- While moving in the period, there is the constant addition of electrons in the nucleus. The shell sie remains constant while electron pull increases from the nucleus, this leads to a reduction in the size of the atom. As the size decreases, it is difficult to remove the electron from the atom, and thus IE value increases in the case of period 2.
- When we study the case of period 4, there is an introduction of d-electrons. As the inner shell electron increases, there is an increase in the shielding effect. This shielding effect tends to decrease the nuclear attraction between the nucleus and outermost electrons. Ultimately this decreases the IE value in the fourth period. Such a phenomenon is absent in the case of group 2 elements.
- If we speak in terms of orbital energy, the IE value decreases while moving from top to bottom in the period. This is due to the fact that, as we go down in the periodic table, the number of shells increases, and the outermost electron is too far from the nuclear attraction, therefore it can be ejected out easily. This marks a decrease in IE value.
To learn more about ionization refer the link:
brainly.com/question/1558319
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7.70 x 10 to the power of 22 molecules