The question is incomplete, the complete question is:
The element tin has the following number of electrons per shell: 2.8. 18, 18, 4. Notice that the number of electrons in the outer shell of a tin atom is the same as that for a carbon atom. Therefore, what must be true of tin? Tin is a polar atom and can bind to other polar atoms. Tin has a high molecular weight to give tin-containing molecules greater stabilty. All of the above Tin conform single covalent bonds with other elements, but not double or triple covalent bonds Tincan bind to up to four elements at a time
Answer:
Tin can bind to up to four elements at a time
Explanation:
Certain important points were made in the question about tin and one of them is that tin is an element in the same group as carbon hence it has the same number of valence electrons as carbon.
Carbon is always tetra valent. The tetra valency of carbon is the idea that carbon forms four bonds.
If tin has the same number of valence electrons as carbon, then, tin can bind to up to four elements at a time
Answer: Strictly a laboratory analysis and can only be done using the data obtained during analysis
Explanation:
To find a solution to this problem, you need to use the data collected during the lab work. A guide could be finding the possible forms of hydrated copper chlorides in reference books. Since it's also a lab work, you can definitely compare your data with lab mates.
The formula CuxCly.zH₂O and its name chloride hydrate already gives you an idea of the possibilities of the value of the integers, hence you can take a good guess for the identity of the unknown salt and calculate the theoretical formular weight for it. From the that you can proceed to also find the mass of water and copper from your lab analysis.
We can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹
Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
Answer:
=60 milligrams
Explanation:
12 x 5
=60 milligrams
Have a nice day!!!!!!! :-)
<u>KA</u>
<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
