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belka [17]
3 years ago
8

How many atoms are in 169g of calcium?

Chemistry
1 answer:
uysha [10]3 years ago
5 0
The answer is 6.022• 10^23 atoms

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a coiled spring

Explanation:

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2 years ago
A tone that is lower in pitch is lower in what characteristic?
zepelin [54]
A tone that is lower in pitch is lower in what characteristic?=frequency
4 0
3 years ago
Read 2 more answers
What is 30-9x2*2-21-4+4
IgorC [24]
Answer:
9 (1-2x²)

Explanation:
The given expression is:
30 - 9x²*2 - 21 - 4 + 4
The first step is to compute the multiplication. This will give:
30 - 18x² - 21 - 4 + 4
Then, we will add like terms as follows:
(30-21-4+4) - 18x²
= 9 - 18x²
Finally, we can take the 9 as a common factor from both terms, this will give:
9 (1-2x²)

Hope this helps :)

5 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

4 0
3 years ago
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