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3 years ago

How do the names of covalent compounds differ from the names of ionic compounds?

1 answer:

5 0

Covalent bonds are named by using prefixes that give the number of atoms of each element in the compound. Ionic compounds are named by replacing the ending of the nonmetal with ate or ite. 6.

plz brain list me 5 star me thank me but thank you

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Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energ

Ahat [919]

Answer: The Gibbs free energy of the reaction is 21.32 kJ/mol

Explanation:

The chemical equation follows:

$\text{Malate }+NAD^+\rightleftharpoons \text{Oxaloacetate }+NADH$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln K_{eq}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = standard Gibbs free energy = 29.7 kJ/mol = 29700 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = $37^oC=[273+37]K=310K$

$K_{eq}$ = Ratio of concentration of products and reactants = $\frac{\text{[Oxaloacetate]}[NADH]}{\text{[Malate]}[NAD^+]}$

$\text{[Oxaloacetate]}=0.130mM$

$[NADH]=2.0\times 10^2mM$

$\text{[Malate]}=1.37mM$

$[NAD^+]=490mM$

Putting values in above expression, we get:

$\Delta G=29700J/mol+(8.314J/K.mol\times 310K\times \ln (\frac{0.130\times 2.0\times 10^2}{1.37\times 490}))\\\\\Delta G=21320.7J/mol=21.32kJ/mol$

Hence, the Gibbs free energy of the reaction is 21.32 kJ/mol

4 0

3 years ago

Why is the melting point of a network solid so high? How is this related to what a network solid is?

Alexxandr [17]

I believe that the answer to the question provided above is that molecules in solid are compact and are connected. It requires huge energy to move it, thus requires high temperature.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

4 0

3 years ago

A radioactive isotope X has a half-life of 36 years. How much X would remain after 108 years, if we started with 500 grams? A. 2

givi [52]

C. 62.5 I guess lol I hate chemistry

4 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago

Given the chemical solution: Na2CO3 + AgNO3 will a precipitate be formed?

True [87]

First we have to establish the chemical reaction. As a result of adding Na2CO3 and AgNO3 products formed are NaNO3 and Ag2CO3. NaNO3 is a soluble substance when dissolved in water.
As a general rule, most Ag, Pb and Hg(I) compounds are considered insoluble. Ag2CO3 is insoluble in water hence the reaction given forms a precipitate.

7 0

4 years ago

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