Remark
The given thing on the right is a positron. The mass for these subatomic particles is considered to be 0. It's atomic number is 1 which means it is a blood relative of a proton.
So essentially what happens is that X is one space to the left on the periodic table. But let's solve this a little bit more formally.
Solution
y stays the same at 147. It is z that changes.
65 = z + 1 Subtract 1 from both sides.
64 = z
So the chemical with 64 as its position on the periodic table is
Gadolinium and the answer is C
Answer:
HCl (aq) + KOH (aq) --------> KCl (aq) + H2O (l) [balanced molecular equation]
Explanation:
Balanced molecular equation has all the components of reaction written as "molecules" thus the name molecular (the other equations break appropriate components into ions, those are total ionic or net ionic equation)
We know hydrochloride acid solution is written as HCl (aq) and potassium hydroxide solution is written as KOH (aq). We know from our knowledge that when acids react with bases they make salt and water, so now we react!
HCl (aq) + KOH (aq) --------> KCl (aq) + H2O (l) [balanced molecular equation]
If you need half the volume, you need double the pressure (V * p). Normal air pressure in the atmosphere is 101300 Pa so that means you need double that, thus 202600 Pa (or 2,026 bars)
Answer:
37.25 grams/L.
Explanation:
- Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.
<em>M = (no. of moles of KCl)/(volume of the solution (L))</em>
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∵ no. of moles of KCl = (mass of KCl)/(molar mass of KCl)
∴ M = [(mass of KCl)/(molar mass of KCl)]/(volume of the solution (L))
∴ (mass of KCl)/(volume of the solution (L)) = (M)*(molar mass of KCl) = (0.5 M)*(74.5 g/mol) = 37.25 g/L.
<em>So, the grams/L of KCl = 37.25 grams/L.</em>
Answer:
= 20.82 g of BaCl2
Explanation:
Given,
Volume = 200 mL
Molarity = 0.500 M
Therefore;
Moles = molarity × volume
= 0.2 L × 0.5 M
= 0.1 mole
But; molar mass of BaCl2 is 208.236 g/mole
Therefore; 0.1 mole of BaCl2 will be equivalent to;
= 208.236 g/mol x 0.1 mol
= 20.82 g
Therefore, the mass of BaCl2 in grams required is 20.82 g
