Remember that a cation will be smaller than its neutral atom, and an anion will be larger than its neutral atom. This would automatically eliminate answer choices A and D.
Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up.
Atomic radii increases as you move from right to left and as you go from up to down.
As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.
The correct answer is B. Br
<h3><u>Answer;</u></h3>
Molarity = 0.25 M
<h3><u>Explanation;</u></h3>
Molarity is given by moles/Liter.
First we find moles:
Number of moles = Mass /molar mass
= (10.7g NH4Cl)/(53.5g/mol NH4Cl)
= 0.200 moles NH4Cl
Then we convert to liters:
= (800mL)*(1L/1000mL) = 0.800L
Therefore; molarity = 0.2moles/0.8L
= 0.25M
Answer:
ΔG° of reaction = -47.3 x 
J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x 
M
[ATP] = 1.2 x 
M
[ADP] = 8.4 x M
Let's plug in these values in the above equation for equilibrium constant:
K = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
K = 1.47 x M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5 ) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x ) + (-16810.68)
ΔG° of reaction = -47.3 x J/mol
The volume of nitrogen monoxide that occupy at STP is= 277 Ml
calculation
The volume is obtained using the combined gas law that is P1V1/T1= P2V2?
Where P1 = 720 MmHg
V1 = 400ml
T1= 100 +273= 373 K
At STP temperature = 273 K and pressure= 760 mm hg
therefore T2= 273 k
p2 = 760 mmhg
V2=?
make V2 the subject of the formula
V2= (T2 ×P1 ×V1)/(P2×T1)
V2 is therefore = (273k x720 mmhg x 400 ml)/(760 mmhg x373K) = 277 Ml
We're given the [OH⁻] as 8.34 × 10⁻¹² M. Using the formula pOH = -log[OH⁻], the pOH of this solution would be -log(8.34 × 10⁻¹²) ≈ 11.08.
The pOH is, for lack of a better term, the "opposite" of pH: A pOH of 7 is neutral; a pOH less than 7 is <em>basic</em>; and a pOH greater than 7 is <em>acidic</em>.
This follows from the relation, pH + pOH = 14. In this case, with a pOH of 11.08, our pH would be 14 - 11.08 = 2.92, which is acidic (pH < 7).
Thus, the correct answer choice is B.
