Answer:
²⁵⁰₉₆Cm → ²⁴⁶₉₆Cm + 4 ¹₀n
Explanation:
The complete equation is;
²⁵⁰₉₆Cm → ²⁴⁶₉₆Cm + 4 ¹₀n
- The above equation is an example of a nuclear reaction in which unstable atom of Cm emits neutrons to become more stable.
- Radioactive isotopes undergo radioactivity or decay to attain stability, they do so by emitting particles such as alpha, beta particles or a neutron.
- An atom of Cm-250 undergoes decay and emits four neutrons to form an atom of Cm-246.
Answer:
1.047 M
Explanation:
The given reaction:
For dichromate :
Molarity = 0.254 M
Volume = 15.8 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 15.8 ×10⁻³ L
Thus, moles of dichromate :
Moles of dichromate = 0.0040132 moles
1 mole of dichromate react with 6 moles of iron(II) solution
Thus,
0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution
Moles of iron(II) solution = 0.02408 moles
Volume = 23 mL = 0.023 L
Considering:
<u>Molarity = 0.02408 / 0.023 = 1.047 M</u>
Answer:
i believe the answer is Aluminum (Al)
Explanation:
Each element has an equal amount of protons to electron so i you subtract one of either you will have to subtract from the other to make sure it is stable. aluminum has 13 protons and electrons. Subtracting 3 from it would gkve it the same qualities as Neon.
Answer:
The density of the liquid in beaker B is less than the that of ice.
Explanation:
Ice will float if its mass is less than the mass of the liquid it displaces.
For example, the density of ice is less than that of water.
A 10 cm³ cube of ice has a mass of about 9 g, while the mass of 10 cm³ of water is 10 g. Thus, 9 g of ice displaces 10 g of water.
The denser water displaces the lighter ice and the ice floats to the top.
If the density of the liquid is <em>less than</em> that of water, say, 8 g/cm³, the ice will displace only 8 g of the liquid. The ice will sink.
Answer:
Rate = 43 M⁻¹s⁻¹[NO₂][O₃]
Explanation:
We need to find the reaction order in
rate = (NO₂ )ᵃ (O₃ )ᵇ
given:
( NO₂ ) M ( O₃ ) M Rate M/s
0.10 0.33 1.420 (1)
0.10 0.66 2.840 (2)
0.25 0.66 7.10 (3)
When keeping the NO₂ concentration constant in the first two while doubling the concentration of O₃ , the rate doubles. Therefore it is first order with respect to O₃
Comparing (2) and (3) increasing the concentration of NO₂ by a factor of 2.5 and keeping O₃ constant , increased the rate by a factor of 2.5. Therefore the rate is first order with respect to NO₂
Then rate law is
= k (NO₂) (O₃ )
To find k take any of the three and substitute the values to find k:
1.420 M/s = k (0.10)M x (0.33)M ⇒ k = 43 /Ms
Then the answer is Rate = 43 M⁻¹s⁻¹[NO₂][O₃]
