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2 years ago

WILL GIVE BRAINLYST

Chemistry

1 answer:

Andrew [12]2 years ago

3 0

Answer: 12.18 u

Explanation: The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its stable isotopes.

In other words, each stable isotope will contribute to the average mass of the element proportionally to its abundance.

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What is the pH of a 0.00530 M solution of HCI?

MrMuchimi

Answer:

2.28

Explanation:

HCl(l) ===> H+ + cl-

HCl is a very strong acid. Almost all of it will decompose to the right. That means the concentration of H+ is 0.00530

pH = - log [H+]

pH = - log[0.00530]

pH = - - 2.2757

pH = 2.2757

Rounded this 2.28

5 0

3 years ago

If Log 4 (x) = 12, then log 2 (x / 4) is equal to

Alexus [3.1K]

The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

= log₂(2²⁴/2²)

= log₂(2²⁴ ⁻ ²)

= log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

#SPJ9

8 0

1 year ago

Which of the following are considered pure substances?

Helen [10]

B.Elements

Explanation: they cannot be separated

8 0

3 years ago

Gold has a molar mass of 197 g/mol. (a) how many moles of gold are in a 3.98 g sample of pure gold? (b) how many atoms are in th

Natali5045456 [20]

A.)49.4974874 moles or 49.5 moles
B.)2.980808730172671e+25 or 3e+25

6 0

3 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago