Answers
Answer:
No additional particle was produced during the decay.
Explanation:
The equation of decay is given as;
¹⁰₆C + ⁰₋₁ e → ¹⁰₅B + x
To identify x, we have to calculate its atomic and mass number.
In the reactants side;
Atomic Number = 6 + (-1) = 5
Mass number = 10 + 0 = 10
In the products side;
Atomic Number = 5 + x
Mass Number = 10 + x
Generally, reactant = product
Atomic Number;
5 = 5 + x
x = 5 - 5 = 0
Mass Number;
10 = 10 + x
x = 10 - 10 = 0
This means no additional particle was produced during the decay.
Related Questions
How would you explain the path of alpha particles in relation to each subatomic particle in the plum pudding model?
Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)? A. NO (aq) B. 2Ba (aq) C. 2Ag (aq) D. CI(aq)
A student is heating a chemical in a beaker with a Bunsen burner.In a paragraph of at least 150 words, identify the safety equipment that should be used and the purpose of it for the given scenario.
What is the daughter nucleus produced when 63 Zn undergoes electron capture? Replace each question mark with the appropriate integer or symbol.
b. Evaporate minerals are more soluble than calcite and quartz.
c. Evaporate minerals are more soluble than quartz and less soluble than calcite.
d. Evaporate minerals are less soluble than quartz and calcite.
Answers
Answer:
Evaporate minerals are more soluble than calcite and quartz.
Explanation:
Evaporate minerals are the water soluble minerals which at higher concentration precipitate out and crystallized forming rocks.
example of chemicals present are:
chlorides and sulphates.
Quartz is silica (very less soluble, or insoluble)
Calcite is calcium carbonate, again an insoluble salt.
Thus
Evaporate minerals are more soluble than calcite and quartz.
Answers
Answers
The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
What is percentage mass?
The percentage mass is the ratio of the mass of the element or molecule in the given compound.
The percentage can be given as:
The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.
Thus,
Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.
Learn more about percentage mass:
Answer:
10.6%
Explanation:
The determined percent mass of water can be calculated from the formula of the hydrate by
dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiplying this fraction by 100.
Manganese(ii) sulphate monohydrate is MnSO4 . H2O
1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of
hydration must be included.
1 Manganes 52.94 g = 63.55 g
1 Sulphur 32.07 g =
32.07 g 2 Hydrogen is = 2.02 g
4 Oygen =
64.00 g 1 Oxygen 16.00 = 16.00 g
151.01 g/mol 18.02 g/mol
Formula Mass = 151.01 + (18.02) = 169.03 g/mol
2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and
multiply this fraction by 100.
Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%
The final result is 10.6% after the two steps calculations
Answers
liquid 1 and 2 have the same color and mass so the answer would be liquid 1 and 2
Explanation:
hope this is helpful
Answers
Answer:
chongus because he's the only good one
i) Before adding NaOH
ii) After adding 24.00 mL NaOH
Answers
Answer:
i) pH = 0.6990
ii) pH = 2.389
Explanation:
i) Before adding aqueous NaOH, there are 25.00 mL of 0.2000 M HCl. HCl reacts with the water in the aqueous solution as follows:
HCl + H₂O ⇒ H₃O⁺ + Cl⁻
The HCl and H₃O⁺ are related to each other through a 1:1 molar ratio, so the concentration of H₃O⁺ is equal to the HCl concentration.
The pH is related to the hydronium ion concentration as follows:
pH = -log([H₃O⁺]) = -log(0.2000) = 0.699
ii) Addition of NaOH causes the following reaction:
H₃O⁺ + NaOH ⇒ 2H₂O + Na⁺
The H₃O⁺ and NaOH react in a 1:1 molar ratio. The amount of NaOH added is calculated:
n = CV = (0.2000 mol/L)(24.00 mL) = 4.800 mmol NaOH
Thus, 4.800 mmol of H₃O⁺ were neutralized.
The initial amount of H₃O⁺ present was:
n = CV = (0.2000 mol/L)(25.00 mL) = 5.000 mmol H₃O⁺
The amount of H₃O⁺ that remains after addition of NaOH is:
(5.000 mmol) - (4.800 mmol) = 0.2000 mmol
The concentration of H₃O⁺ is the amount of H₃O⁺ divided by the total volume. The total volume is (25.00 mL) + (24.00 mL ) = 49.00 mL
C = n/V = (0.2000 mmol) / (49.00 mL) = 0.004082 M
The pH is finally calculated:
pH = -log([H₃O⁺]) = -log(0.004082) = 2.389