thx but it's actually 143
Answer:
The mass of the element is 26.20 amu
Explanation:
In this question, we are asked to calculate the mass of an element with 15 protons, 13 electrons and 11 neutrons
To calculate the atomic mass of the element, we take into consideration the masses of the individual sub atomic particles
Electrons have 0 atomic mass unit(their masses are negligible) we have no business here, Protons have a mass of
1.00727647 amu, while the mass of neutron is 1.0086654 amu
The mass of 15 protons is thus 15 * 1.00727647 = 15.10914705 amu
The mass of 11 neutrons is 11 * 1.0086654 =
11.0953194 amu
Adding this together, we have ; 11.0953194 + 15.10914705 = approximately 26.20 amu
Answer:
See answer below
Explanation:
In this case, let's draw the butane molecule:
CH₃ - CH₂ - CH₂ - CH₃
According to what the exercise states, we removed an atom of hydrogen from the frist carbon. This could be any of the terminals. I'll grab the first from left to right.
CH₂⁺ - CH₂ - CH₂ - CH₃
When this happens, the atom of carbon is lacking one space and it forms a carbocation.
Followed this step, an hydroxile group replace the atom of hydrogen. The hydroxile is the OH, and when we have an alkane with an OH group in the molecule, we are actually converting this molecule into an alcohol, therefore the molecule formed is:
<h2>
OH - CH₂ - CH₂ - CH₂ - CH₃</h2><h2 />
Hope this helps
M = GFM × N
Sodium Oxide
s Na O
v 1 2
s 2 1
d 2 1
f Na2O
Na2O
2 × 23 = 46
1 × 16 = 16
46 + 16 = 62g
GFM = 62g
N = 2.24 moles
M = ?
M = GFM × N
M = 62,2 × 2.24
M = 138.88g
Answer:
0.542M HCl
Explanation:
The reaction of H₂SO₄ with NaOH is:
H₂SO₄ + 2 NaOH → 2H₂O + Na₂SO₄
<em>Where 1 mole of acid reacts with 2 moles of NaOH</em>
Moles of H₂SO₄ are:
0.0500L × (0.361mol / L) = 0.01805 moles H₂SO₄
Thus, moles of NaOH that neutralize this acid are:
0.01805 moles H₂SO₄ × (2 mol NaOH / 1 mol H₂SO₄) = 0.0361 moles NaOH
And concentration is:
0.0361 moles NaOH / 0.0200L = <em>1.805M</em>
And, reaction of NaOH with HCl is:
NaOH + HCl → H₂O + NaCl
<em>Where 1 mole of NaOH reacts per mole of NaOH</em>
As you use 30.0mL = 0.0300L of NaOH to neutralize the HCl acid, moles of acid are:
0.0300L × (1.805mol / L) = 0.05415 moles NaOH = moles HCl
In 0.1000L:
0.05415 moles HCl / 0.1000L = <em>0.542M HCl</em>
