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3 years ago

To begin legal proceedings against

Chemistry

1 answer:

Rudik [331]3 years ago

8 0

What is the question? :)

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How many moles are in 12 grams of NH3?

aleksklad [387]

Answer:

molecular weight of NH3 or grams This compound is also known as Ammonia. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles NH3, or 17.03052 grams.

8 0

3 years ago

Pyridine, C5H5N, is a polar organic solvent. How many carbon atoms are in 7.05 moles of pyridine? 4.24 x 1024 4.24 x 10, 24 2.12

leonid [27]

Answer:

2.122×10^25atoms

Explanation:

number of moles=mass/molar mass

7.05moles= mass of pyridine/79

reacting mass of pyridine=556.95

C5H5N= (12×5)+(5)+(14)=79

C5=60

to find the mass of carbon in 556.95g of pyridine we take the stoichometric ratio

60[C5] -----> 79[C5H5N]

x[C5] --------> 556.95g[C5H5N]

cross multiply

x=(60×556.95)/79

x=423g of carbon

moles=mass/molar mass

moles of carbon=423/12

moles=35.25moles of carbon

moles=number of particles/Avogadro's constant

35.25=number of particles/6.02×10^23

number of particles=2.122×10^25atoms of carbon

3 0

2 years ago

A sample of octane (c8h18 that has a mass of 0.750 g is burned in a bomb calorimeter. as a result, the temperature of the calori

timurjin [86]

In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:

Heat of the combustion reaction = mC(T2-T1)
= 1 (1.50) (41-21)
= 30 kJ

7 0

3 years ago

Read 2 more answers

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$