Answer:
The answer of this question is molecule
Answer:
459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution
Explanation:
Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by the following expression:
Molarity is expressed in units 
In this case:
- Molarity: 1.56 M= 1.56
- Number of moles of calcium chlorine= ?
- Volume= 2.657 liters
Replacing:
Solving:
Number of moles of calcium chlorine= 1.56 M* 2.657 liters
Number of moles of calcium chlorine= 4.14 moles
In other side, you know:
- Ca: 40 g/mole
- Cl: 35.45 g/mole
Then the molar mass of the calcium chloride CaCl₂ is:
CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole
Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?
mass= 459.126 g
<u><em>459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution</em></u>
Answer:
0.241 M
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
HBr + NaOH —> NaBr + H₂O
From the balanced equation above,
The mole ratio of acid, HBr (nₐ) = 1
The mole ratio of base, NaOH (n₆) = 1
Finally, we shall determine the concentration of the NaOH solution. This can be obtained as follow:
Volume of base, NaOH (V₆) = 20 mL
Volume of acid, HBr (Vₐ) = 24.1 mL
Concentration of acid, HBr (Cₐ) = 0.2 M
Concentration of base, NaOH (C₆) =?
CₐVₐ / C₆V₆ = nₐ/n₆
0.2 × 24.1 / C₆ × 20 = 1/1
4.82 / C₆ × 20 = 1
Cross multiply
C₆ × 20 = 4.82
Divide both side by 20
C₆ = 4.82 / 20
C₆ = 0.241 M
Therefore, the concentration of the NaOH solution is 0.241 M
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
Answer:1.
Explanation: This reaction is catalyzes by pyruvate dehydrogenase. Pyruvate being the end product of glycolysis has many fates after glycolysis,one of which is to enter the TCA(Tricarboxylic acid cycle) cycle. It is first converted to actetate by the action of pyruvate dehydrogenase. This enzyme converts pyruvate to acetate releasing CO2 and NADH because this oxidative decarboxylation of pyruvate is coupled with reduction of NAD+ which can feed into the electron transport chain.
