If a neutral atom has 5 protons (+) and 5 neutrons (0), how many electrons does it have?

In a redox reaction, how does the total number of electrons lost by the oxidized substance compare to the total number of electr

snow_lady [41]

The answer is (2) equal to. In redox reactions, you can't just lose electrons somewhere. If an electrons is lost by one, it must be gained by another. Hence, the importance of balancing redox reactions.

3 0

3 years ago

How much energy is needed to change 44 grams of ice at -8° to steam at 107°?

Tpy6a [65]

Q=m(c∆t +heat of fusion + heat of evaporation)

m= 44g
c= 4.186 J/g.C
∆t= 107-(-8) =115 C
heat of fusion= 333.55 J/g
heat of evaporation=2260 J/g

Q=44(4.186*115 + 333.55 + 2260)
Q= 135297.36 J

5 0

3 years ago

Cooking an egg is an example of what type of change

Katen [24]

Answer: chemical change

7 0

2 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago

Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y

d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = $10^{-3.5}$ = $3.162 * 10^{-4}$

From the Ice table

$3.162 * 10^{-4}$ = $\frac{x + H^+}{[aspirin]}$ = $\frac{x^{2} }{0.012-x}$

given that the value of Ka is small we will ignore -x

x² = $3.162 * 10^{-4} * 0.012$

x = $1.948 * 10^{-3}$

Therefore

[ H⁺ ] = $1.948 * 10^{-3}$

given that

pH = - Log [ H⁺ ]

= - ( -3 + log 1.948 )

= 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0

2 years ago