Which cell copies its dna

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat

Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:

N: 14 g/mol
H: 1 g/mol
O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
O₂: 2*16 g/mol= 32 g/mol
NO: 14 g/mol + 16 g/mol= 30 g/mol
H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

NH₃: 4 moles*17 g/mol= 68 g
O₂: 5 moles*32 g/mol= 160 g
NO: 4 moles*30 g/mol= 120 g
H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

mass of O₂≅17.65 g

17.65 grams of O2 are needed for a complete reaction.

3 0

3 years ago

Question 2 (1 point) Al + S8 = Al2S3 When balanced, what is the coefficient of Al?

gtnhenbr [62]

Answer:

16

Explanation:

FIRST AND FOREMOST, BALANCE YOUR EQUATION.

Al + S8 ➡️ Al2S3

Numbers of Al=1. ➡️ Numbers of Al = 2

Numbers of S =8. ➡️ Numbers of S. = 3

USE COEFFICIENT TO BALANCE THE EQUATION.

16Al + 3S8 ➡️ 8Al2S3

Now the Numbers of Al and S in both sides of eqn. is balanced

The Answer Is 16

3 0

3 years ago

Read 2 more answers

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the

IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Explanation :

The expression for first order reaction is:

$[C_t]=[C_o]e^{-kt}$

where,

$[C_t]$ = concentration at time 't' (final) = ?

$[C_o]$ = concentration at time '0' (initial) = 0.100 M

k = rate constant = $5.40\times 10^{-3}s^{-1}$

t = time = 17.0 min = 1020 s (1 min = 60 s)

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

$[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}$

$[C_t]=4.05\times 10^{-4}M$

Thus, the concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

4 0

3 years ago

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th

Vera_Pavlovna [14]

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ( $HN_3$ ). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$ . So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$