Which of the following is the correct Lewis structure for carbon?

What happens first during cellular respiration

Yuliya22 [10]

Answer:

glcolysis

Explanation:

Glcolysis uses glucose to create ATP within the cytoplasm of the cell.

7 0

3 years ago

What is the electron configuration of a noble gas?

Debora [2.8K]

Answer:

A noble gas configuration of an atom consists of the elemental symbol of the last noble gas prior to that atom, followed by the configuration of the remaining electrons. So for sodium, we make the substitution of [Ne] for the 1s22s22p6 part of the configuration. Sodium's noble gas configuration becomes [Ne]3s1. That is the answer.

Explanation:

Brainliest Please.

5 0

3 years ago

The Henry's law constant (kH) for O2 in water at 20°C is 1.28e-3 mol/l atm. How many grams of O2 will dissolve in 3.5 L of H2O t

Sophie [7]

Answer : The mass of $O_2$ dissolved will be, 0.2365 grams

Explanation :

First we have to calculate the concentration of $O_2$ .

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = concentration of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 1.65 atm

$k_H$ = Henry's law constant = $1.28\times 10^{-3}mole/L.atm$

Now put all the given values in the above formula, we get:

$C_{O_2}=(1.28\times 10^{-3}mole/L.atm)\times (1.65atm)$

$C_{O_2}=2.112\times 10^{-3}mole/L$

The concentration of $O_2$ = $2.112\times 10^{-3}mole/L$

Now we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\text{Concentration of }O_2\times \text{volume of solution}$

$\text{Moles of }O_2=(2.112\times 10^{-3}mole/L)\times (3.5L)=7.392\times 10^{-3}mole$

Now we have to calculate the mass of $O_2$

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(7.392\times 10^{-3}mole)\times (32g/mole)=0.2365g$

Therefore, the mass of $O_2$ dissolved will be, 0.2365 grams

4 0

3 years ago

The formula in my book for carbonate is Co32- and I’m confused because carbon has 4 valence electrons and oxygen has 6 . How doe

balandron [24]

Hello!!!

what exactly are you wanting to know? It isn’t clear from your question. Please advise and I will answer it for you ASAP. I think I know what you are wanting but I don’t want to assume.... thank you

4 0

3 years ago

Read 2 more answers

If 2.00g of p-aminophenol ( 109.1 g/mol) reacts with 5.00 ml of acetic anhydride (102.1 g/mol and density = 1.08 g/ml), what mas

inna [77]

The complete reaction is as,

4-Aminophenol + Acetic Anhydride → Acetaminophen + Acetic Acid

First of all convert the ml of Acetic anhydrite to grams,
As,
Density = mass / volume
Solving for mass,
mass = Density × Volume
Putting values,
mass = 1.08 g/ml × 5ml

mass = 5.4 g of acetic anhydride

First Find amount of acetic anhydride required to react completely with 2 g of p-Aminophenol,
As,
109.1 g of p-aminophenol required = 102.1 g of acetic anhydride
so, 2 g of p-aminophenol will require = X g of Acetic Anhydride

Solving for X,
X = (2 g × 102.1 g) ÷ 109.1 g

X = 1.87 g of acetic anhydride is required to be reacted.

But, we are provided with 5.4 g of Acetic Anhydride, means p-aminophenol is the limiting reactant and it controls the formation of product. Now Let's calculate for product,
As,
109.1 g of p-aminophenol produced = 180.2 g of Acetaminophen
So 2.00 g of p-aminophenol will produce = X g of Acetaminophen

Solving for X,
X = (2.00 g × 180.2 g) ÷ 109.1 g

X = 3.30 g of Acetaminophen

Result:
If 2.00g of p-aminophenol reacts with 5.00 ml of acetic anhydride 3.30 g of acetaminophen is made.

4 0

3 years ago