How many moles of NaOH are present in 65.00 g of NaOH(s)? *

Chemistry

1 answer:

schepotkina [342]2 years ago

6 0

Answer:

<h2>The answer is option B</h2>

Explanation:

To find the number of moles we must first find the molar mass of the compound after that we use the formula

$n = \frac{m}{M}$

where

n is the number of moles

M is the molar mass

m is the mass

M(Na) = 23 , M(O) = 16 , M(H) = 1

Molar mass ( NaOH ) = 23+ 16 + 1

= 40 g/mol

mass = 65 g

Substitute the values into the above formula and solve

That's

$n = \frac{65}{40} = \frac{13}{8}$

We have the final answer as

<h3>1.625 moles </h3>

Hope this helps you

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In the citric acid cycle, after the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C

Alborosie

Answer:

The carbons of the acetyl group oxidize which generate CO2, and in turn H2O.

Explanation:

The pyruvic acid that is generated during glycolysis enters the mitochondria. Inside this organelle, the acid molecules undergo a process called oxidative decaborxylation in which an enzyme of several cofactors is involved, one of which is coenzyme A. Pyruvic acid is transformed into an acetyl molecule and these are been introduced to the begining of the Krebs Cycle where the acetyl-group (2C) from acetyl-CoA is transferred to oxaloacetate (4C) to produce citrate (6C). As the molecule cycles the two carbons of the acetyl oxidize and are released in the form of CO2. Then the energy of the Krebs cycle becomes sufficient to reduce three NAD +, which means that three NADH molecules are formed. Although a small portion of energy is used to generate ATP, most of it is used to reduce not only the NAD + but also the FAD which, if oxidized, passes to its reduced state, FADH2

8 0

2 years ago

Anyway, which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

marissa [1.9K]

Nitrate ions $\text{NO}_3^{-}$ and potassium ions $\text{K}^{+}$ .

<h3>Explanation</h3>

Potassium dichromate undergoes a double-decomposition reaction with barium nitrate to produce barium dichromate and potassium nitrate. The reaction is possible due to the low solubility of barium dichromate that precipitate out of the solution readily after its production.

Start with the balanced chemical equation for this process:

$\text{K}_2\text{CrO}_4 \; (aq) + \text{Ba}(\text{NO}_3)_2 \; (aq) \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{KNO}_3 \; (aq)$

Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

$2 \; \text{K}^{+}\; (aq)+ 2\; \text{CrO}_4^{-} \; (aq) + \text{Ba}^{2+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq) \\ \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{K}^{+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq)$

$\text{K}^{+}$ and $\text{CrO}_4^{-}$ are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.