Which molecule among these is polar? NH3 BCl3 O2 NO3

Given the balanced equation 2C4H10 + 13O2 → 8CO2 + 10H2O, how many moles of CO2 are produced when 14.9g of O2 are used?

Anna [14]

Answer: The number of moles of $CO_2$ produced are, 0.287 moles.

Explanation : Given,

Mass of $O_2$ = 14.9 g

Molar mass of $O_2$ = 32 g/mol

First we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}$

$\text{Moles of }O_2=\frac{14.9g}{32g/mol}=0.466mol$

Now we have to calculate the moles of $CO_2$

The balanced chemical equation is:

$2C_4H_{10}+13O_2\rightarrow 10H_2O+8CO_2$

From the reaction, we conclude that

As, 13 mole of $O_2$ react to give 8 moles of $CO_2$

So, 0.466 mole of $O_2$ react to give $\frac{8}{13}\times 0.466=0.287$ mole of $CO_2$

Therefore, the number of moles of $CO_2$ produced are, 0.287 moles.

3 0

3 years ago

PLLLLEEAASEEE HELLPP!!

nekit [7.7K]

Answer: The correct option is C) 1.68 mol/L

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (L)}}$ .....(1)

Given values:

Given mass of $C_6H_{12}O_6$ = 150 g

Molar mass of $C_6H_{12}O_6$ = 180 g/mol

Volume of the solution = 0.50 L

Putting values in equation 1, we get:

$\text{Molarity of solution}=\frac{150}{180\times 0.50}\\\\\text{Molarity of solution}=1.68M$

Hence, the correct option is C) 1.68 mol/L

7 0

3 years ago

What is the reaction rate for a chemical reaction?

lakkis [162]

Some are essentially instantaneous, while others may take years to reach equilibrium. The Reaction Rate for a given chemical reaction is the measure of the change in concentration of the reactants or the change in concentration of the products per unit time.

8 0

3 years ago

Read 2 more answers

What is happening in the diagram?

OlgaM077 [116]

Answer:It is a picture with a boat sailing on the water with sound waves going down to an old and sunken ship but the waves then go back up to the boat in a different angle.

Explanation:

hope it works thx

4 0

4 years ago

An unknown acid was titrated two times with 0.10 M NaOH. The first titration was done using an indicator to determine the equiva

Anit [1.1K]

Answer:

hello your question is incomplete attached below is the complete question

a) 0.12 M

b) Ka = 3.0 * 10^-7

c) Alizarin Yellow R

Explanation:

A) Determine the concentration of the unknown acid

The PH of the unknown acid before addition of NaOH is ; 3.72 ( weak acid )

First determine the moles of of NaOH

= molarity * volume

= 0.10 M * 30.0 ML = 0.0030 mol

at equivalence point

moles of NaOH = moles of unknown acid = 0.0030 mol

volume of unknown acid = 25.0 mL

Next calculate the concentration of the Acid ( HA )

= moles / volume

= 0.0030 mol / 25.0 mL

= 0.12 M

b) Determine the Ka of the unknown acid

attached below is the detailed solution

Ka = 3.0 * 10^-7

c) The indicator that would be a good choice to use in the titration of this acid with NaOH is ; Alizarin Yellow R . this is because the titration is between a strong base and a weak acid.

7 0

3 years ago