Answer:
The new volume will be 367mL
Explanation:
Using PV = nRT
V1 = 259mL = 0.000259L
n1 = 0.552moles
At constant temperature and pressure, the value is
P * 0.000259 = 0.552 * RT ------equation 1
= 0.552 / 0.000259
= 2131.274
V2 = ?
n2 = 0.552 + 0.232
n2 = 0.784mole
Using ideal gas equation,
PV = nRT
P * V2 = 0.784 * RT ---------- equation 2
Combining equations 1 and 2 we have;
V2 = 0.784 / 2131.274
V2 = 0.000367L
V2 = 367mL
Rr - Round eyes. the two alleles are R and r, and if R is dominant, the phenotype will be round
Answer:
For any given element, ionization energy increases as subsequent electrons are removed. For example, the energy required to remove an electron from neutral chlorine is 1251 kJ/mol. ... An even sharper increase in ionization energy is witnessed when inner-shell, or core, electrons are removed.
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Answer:
1. the group number of sodium is 1 and it is a metal
2. the group number of helium is 18 and it is a nonmetal
3. the group number of iodine is 17 and it is a nonmetal
4.the group number of calcium is 2 and it is a metal
5. lithium has similar properties to potassium
6. calcium has similar properties to magnesium
7. neon has similar properties to xenon
8. Iodine has similar properties to chlorine
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
