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3 years ago

HELP ASAP PLZ

2 answers:

4 0

D. The total number of atoms when glucose and oxygen react stays the same when carbon dioxide and water are produced. The conservation of the mass is a fundamental law of chemistry and physics. It indicates not only that during any experiment, including if it involves a chemical transformation, the mass is conserved, but also that the number of elements of each chemical species is conserved. Like any law of conservation it is expressed by a conservation equation.

frez [133]3 years ago

3 0

the answer is

D.total number of atoms when glucose and oxygen react stays the same when carbon dioxide and water are produced.

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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:

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

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

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

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

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

5 0

3 years ago

When 1 mol of sodium nitrate, is dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed. What is the drop in

vesna_86 [32]

Answer:

1.0 ° C

Explanation:

The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85

Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃

⇒ 17/85 = 1.38 moles

Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.

when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..

Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.

Using the relation : Q = mcΔT to determine the temperature drop ; we get:

55.2 = 17 × 4 (ΔT)

55.2 = 68 ΔT

ΔT= 0.8 ° C

ΔT ≅ 1.0 ° C

Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C

7 0

3 years ago

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

5 0

3 years ago

What are the 5 common uses of fossil fuels?

Leni [432]

Answer:

Electricity. Coal alone provides half the electricity in the United States. ...

Heating. Oil and natural gas are commonly used for heating homes as well as providing heat for industrial applications.

Transportation. Oil supplies 99 percent of the energy for cars in the form of gasoline and diesel. ...

Limits. ...

Considerations.

Explanation:

5 0

3 years ago

What is Lewis symbol for K+

Alexxx [7]

Lewis symbol for K+ is given by a dot and K.

<h3>Meaning of Lewis symbol</h3>

Lewis symbols can be defined as a symbol the is used in describing the valence electron configurations of an atom ions.

A Lewis symbol consists of the symbol of the element and then surrounded by one dot for each of its valence electrons.

In conclusion, Lewis symbol for K+ is given by a dot and K.

Learn more about Lewis symbol: brainly.com/question/6605265

#SPJ1

7 0

2 years ago