Answer:
hope this helps :)
Explanation:
for the first one, you can look at the periodic table and look at the atomic number and it will show you how many protons there are giving you the answer because protons and electrons are equal in a pure element
a- carbon
b- neon
c- boron
d- oxygen
e- helium
f- hydrogen
g- lithium
h- beryllium
i- nitrogen
1- sulfur
2- S
3- 16
4- 32.066
5- 16
6- 16
7- 16.066
8- draw circles and put 16 dots like on the other page and in the middle put 16 nuetrons and electrons
9- 6 i think
B) the number of electrons
Answer:
2.67 × 10⁻²
Explanation:
Equation for the reaction is expressed as:
CaCrO₄(s) ⇄ Ca₂⁺(aq) + CrO₂⁻⁴(aq)
Given that:
Kc=7.1×10⁻⁴
Kc= ![[Ca^{2+}][CrO^{2-}_4]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5BCrO%5E%7B2-%7D_4%5D)
Kc= [x][x]
Kc= [x²]
7.1×10⁻⁴ = [x²]
x = 
x = 0.0267
x = 
Subsidy
A price support is a type of subsidy. It seems that this is not given in
the choices. However, subsidy is a form of monetary or financial support which
is provided by a local or state sector. It aims to provide economic and social
stability and development. Hence, subsidy encompasses various types of support
funded by either government or non-government offices. It can be given in form
of either direct (money, incentives and etc.) or indirect (insurance,
write-offs and etc.) in procedures.
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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