Answer:
ⁿₐX => ²¹⁸₈₄Po
Explanation:
Let ⁿₐX be the isotope.
Thus, the equation can be written as follow:
²²²₈₆Rn —> ⁴₂α + ⁿₐX
Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:
222 = 4 + n
Collect like terms
222 – 4 = n
218 = n
Thus,
n = 218
86 = 2 + a
Collect like terms
86 – 2 = a
84 = a
Thus,
a = 84
ⁿₐX => ²¹⁸₈₄Po
²²²₈₆Rn —> ⁴₂α + ⁿₐX
²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po
If we convert the ounces to grams, there are approximately 283.495 grams of plant fertiliser
If nitrogen has 15% of this, all we have to do is divide this number by 100 to get the mass of 1% and multiply it by 15.
In the end, we end up with the mass of 42.5243 g
Hope I helped! xx
Answer: The density of the object will be 
Explanation:
Density is defined as the mass contained per unit volume.

Given : Mass of object = 19.6 grams
Volume of object= 
Putting in the values we get:

Thus density of the object will be 
Answer:
HCO₂
Explanation:
From the information given:
The mass of the elements are:
Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g
To determine the empirical formula;
First thing is to find the numbers of moles of each atom.
For Carbon:

For Hydrogen:

For Oxygen:

Now; we use the smallest no of moles to divide the respective moles from above.
For carbon:

For Hydrogen:

For Oxygen:

Thus, the empirical formula is HCO₂
38.46g
Explanation:
Given parameters:
Mass of CaF₂ = 75.0g
Unknown:
Mass of calcium that can be recovered = ?
Solution:
This is a mass percentage problem and we need to solve it accordingly.
To solve this problem;
Find the molar mass of CaF₂
Find the ratio between the molar mass of Ca and that of CaF₂
Multiply by the given mass
Molar mass of CaF₂ = 40 + (2 x 19) = 78g/mol
Mass of calcium =
x 75 = 38.46g
learn more:
Mass percentage brainly.com/question/8170905
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