The mass in grams of each of the following using dimensional analysis is:
- 6.02 x 10²³ atoms of Mg = 12g of Mg
- 3.01 x 10²³ formula units of CaCl2 = 55.5g of CaCl2
- 12.4 x 10¹⁵ molecules of formaldehyde (CH2O) = 0.000000618g of formaldehyde
<h3>How to calculate mass?</h3>
The mass of a substance can be calculated in different ways depending on whether molecules or moles are given.
First, we need to convert atoms/formula units/molecules to moles as follows:
- 6.02 × 10²³ ÷ 6.02 × 10²³ = 1 mole of Mg
- 3.01 x 10²³ ÷ 6.02 × 10²³ = 0.5 moles of CaCl2
- 12.4 x 10¹⁵ ÷ 6.02 × 10²³ = 2.06 × 10-⁸ moles of formaldehyde
Next, we convert each mole value to mass as follows:
- 1 mole of Mg = 1 × 12 = 12g of Mg
- 0.5 moles of CaCl2 = 0.5 × 111 = 55.5g of CaCl2
- 2.06 × 10-⁸ moles of CH2O = 0.000000618g of formaldehyde
Learn more about mass at: brainly.com/question/19694949
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Answer:
the answer is C
Explanation:
This is because group6 elements are diatomic and when they are chemically combined their subscript 2 cancels out
Answer:
1. nitric acid: sa
2. perchloric acid: sa
3. hydrofluoric acid: wa
Explanation:
A strong acid (sa) is the one that is completely dissociated into ions in water. Conversely, a weak acid (wa) is not completely dissociated in water.
From the options, the strong acids are:
1. nitric acid (HNO₃). It dissociates completely into ions when is dissolved in water, as follows:
HNO₃ → H⁺ + NO₃⁻
2. perchloric acid (HClO₄). It is completely dissociated in water as follows:
HClO₄ → H⁺ + ClO₄⁻
The weak acid is hydrofluoric acid (HF). In water, only a small proportion is dissociated into ions. The proportion of ions formed is given by the equilibrium constant Ka. The dissociation is written by using double arrows:
HF ⇄ H⁺ + F⁻
Answer:
Elements only contain one type of atoms while compounds contain two or more types of atoms.
Explanation:
An example of an element is sodium --> Na (only Na atoms)
An example of a compound can be water --> H2O (contain H and O atoms)
*But the particles within a compound are all the same.
-dB/dt = kAB = k(2B)(B) = 2kB^2
-dB/B^2 = 2kdt
Integrating: 1/B - 1/(B_0) = 2kt
At t = 10, if 15 g of C have formed, this must have consumed 10 g A and 5 g B. The remaining mass of B is 45 g.
1/45 - 1/50 = 2(k)(10)
k = 1.11 x 10^-4
Then substituting this value of k with t = 40:
1/B - 1/50 = 2(1.11 x 10^-4)(40)
1/B - 1/50 = 0.008889
1/B = 0.028889
B = 34.62 g remaining
Therefore, 50 - 34.62 = 15.38 g of B have been consumed.
Doubling, 30.76 g of A have been consumed.
This means that 15.38 + 30.76 = 46.15 g of C have been formed.
