Answer:
0.08097 grams of nitrate ions are there in the final solution.
Explanation:
Moles of cobalt(II) nitrate ,n= 
Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L
Let the molarity of the solution be 
A students then takes 4 .00 mL of solution and dilute it to 275 ml.
(molarity after dilution)
(after dilution)
Molarity of the of solution after dilution is 0.002375 M.
1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:
![[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M](https://tex.z-dn.net/?f=%5BNO_3%5E%7B-%7D%5D%3D%5Cfrac%7B2%7D%7B1%7D%5Ctimes%200.002375%20M%3D0.004750%20M)
Moles of nitrate ions = n
Volume of the solution = 275 mL = 0.275 L
Molarity of the nitrate ions = ![[NO_3^{-}]=0.004750 M](https://tex.z-dn.net/?f=%5BNO_3%5E%7B-%7D%5D%3D0.004750%20M)
![[NO_3^{-}]=\frac{n}{0.275 L}](https://tex.z-dn.net/?f=%5BNO_3%5E%7B-%7D%5D%3D%5Cfrac%7Bn%7D%7B0.275%20L%7D)
n = 0.001306 mol
Mass of 0.001306 moles of nitrate ions:
0.001306 mol × 62 g/mol= 0.08097 g
0.08097 grams of nitrate ions are there in the final solution.
a.
Acids react with bases and give salt and water and the products.
Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
To balance the reaction equation, both sides hould have same number of elements.
Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1
Hence, the reaction equation is already balanced.
b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
= 5.681 x 10⁻³ mol
Stoichiometric ratio between NaOH and HCl is 1 : 1
Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol
5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.
Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M
Answer:
Ethynylcyclopropane is the stable isomer for given alkyne.
Explanation:
In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,
CₙH₂ₙ₋₂
Putting value on n = 5,
C₅H₂.₅₋₂
C₅H₈
Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.
C₅H₈ ------------(-2 H) ----------> C₅H₆
Hence, the molecular formula for given compound is C₅H₆
Below, 4 different isomers with molecular formula C₅H₆ are attached.
The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.
Compounds 3-ethylcycloprop-1-yne, <u>cyclopentyne </u>and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.
Answer:
The adjustable legs and the table of sand.
<em>Note:The question is incomplete. The complete question is given below.</em>
Using Models to Answer Questions About Systems
Armando’s class was looking at images of rivers formed by flowing water. Most of the rivers were wide and shallow, but one river was narrow and deep. Armando’s class thinks that this river is narrow and deep because:
- the hill that the water flowed down was very steep, or
- the sand grains that the water flowed through were very small.
Armando designed the model below to try to answer the question: Why is this river so narrow and deep?
Explanation:
The model designed by Armando will be helpful to answer the question because of the following features it possesses:
1. An adjustable leg- since one of the hypotheses put forward by the class to explain why the river was narrow and deep was that the hill that the water flowed down was very steep, the adjustable legs can be lowered or raised in order to make the slope shallower or steeper so that their hypothesis can be tested.
2. A table of sand- the table of sand serves as the streambed. By adjusting the size of the sand grains to be larger or smaller, the students will be able to to test their second hypothesis that the small size sand grains that the water flowed through was the reason for the river to be narrow and deep.
The results of their experiments will enable them to come to a conclusion.
it´s actually Lithium and fluorine / Magnesium and Chlorine / Beryllium and Nitrogen
