Answer:
³⁸₂₀Ca.
Explanation:
³⁸₁₉K –> __ + ⁰₋₁β
Let ʸₓA represent the unknown.
Thus the equation above can be written as:
³⁸₁₉K –> ʸₓA + ⁰₋₁β
Thus, we can obtain the value of y an x as follow:
38 = y + 0
y = 38
19 = x + (–1)
19 = x – 1
Collect like terms
19 + 1 = x
x = 20
Thus,
ʸₓA => ³⁸₂₀A => ³⁸₂₀Ca
Therefore, the equation is:
³⁸₁₉K –> ³⁸₂₀Ca + ⁰₋₁β
Answer : The wavelength is 
Solution : Given,
frequency = 29.2 Hz
Formula used :
where,
= frequency
= wavelength
c = speed of light = 
Now put all the given values in this formula, we get the wavelength.
Therefore, the wavelength is
Answer:
a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) 3.14g must be added
Explanation:
a) For the reaction:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)
As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)
To balance hydrogens, the other coefficients are:
Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:
6.5 g × (1mol / 315.48g) =<em> 0.0206moles of Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:
0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:
0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>
This one is the easiest law, but you would take 53 and 185 and add them together to get 235 and then you will minus 235 and 365 and the answer you are looking for is 130 mmHg! Hopefully this helped you!!
Answer:
Samarium
Explanation:
The element Sm describe is called Samarium. This element has unique sets of properties that makes it very unique and distinct.
The lanthanides are found in the f-block on the periodic table of elements.
This element is a moderately hard silvery metal that readily oxidizes in air. It assumes an oxidation state of +3. The element has an atomic number of 62
