Answer:
atomic number
Explanation:
atomic number is the number of protons
The oxidation state of Chromium chloride (III) is +3
How can we find the answer?
First of all write down what you know about the molecule:
1: The molecule hasn't got an electric charge (is not an ione), this means that either positive and negative charges of its atoms are balanced (we have the same number of positve and negative charges)
2: Since it's a salt, where the metal is chromium and the non metal is the alogen Chlorine, we know that the negative charge belogns to the non metal element because of its elettronegativity, therefore the positive charge belongs to the metale element (chromium).
3: when chlorin forms binary salts its oxidation state is always -1 (you can find out this info in a periodic table)
In <span>CrC<span>l3</span></span> we have 3 chlorine atoms where each of them carrys 1 negative charge, so the total amount of negative charges is -3
Since the charges are balanced, the question is: Which is the positive charge that Chromium must carry in order to balance 3 negative charges?
The answer comes out to +3
It would be F only because I wouldn’t make sense CI doesn’t exhibit crystalline neither does BR so F
Answer:
When iron is heated in air it reacts with oxygen to form the compound iron oxide. Iron is a solid and oxygen is a gas. 5 Draw diagrams to show how the atoms are arranged in iron, oxygen and iron oxide in the circles below their names. Use different colours for the iron atoms and the oxygen atoms.
Explanation:
A) when the balanced equation of the reaction is:
H2CO3(aq) → HCO3 -(aq) + (H+)
and when we have Ka = 4.3 x10^-7 & PH = 7.4
So first we will get PKa = -㏒ Ka
PKa = -㏒(4.3x10^-7) = 6.37 by substitution with Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
PH= PKa + ㏒[HCO3-]/[H2CO3]
㏒[HCO3-]/[H2CO3] = PH-Pka
[HCO3-] /[H2CO3] = 10^(7.4 - 6.37)
∴[HCO3-]/[H2CO3] = 11.7
∴[H2CO3]/[HCO3-] = 1/11.7 = 0.09
B) when The balanced equation for this reaction is:
H2PO42-(aq) → HPO4-(aq) + H+
and when we have Ka = 6.2x10^-8 & PH = 7.15
So Pka= -㏒Ka = -㏒(6.2x10^-8) = 7.2 by substitution by Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
7.15= 7.2 + ㏒[HPO4]/[H2PO4]
-0.05 = ㏒[HPO4]/[H2PO4]
∴[HPO4]/[H2PO4] = 10^-0.05 = 0.89
∴[H2PO4]/[HPO4] = 1/0.89 = 1.12
c) H3PO4(aq) ↔ H2PO-(aq) + H+
the answer is: because we have Ka =7.5x10^-3 and it is a high value of Ka to make a good buffer, also we need a week acid with th salt of the week acid as H3PO4 is a strong acid so it does'nt make a goof buffer.
