<h3>Answer </h3>
After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left. after three half lives or 17190 years, 8.75 g of C-14 will be
Explanation:
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Answer:
50,849.25 Joules
Explanation:
The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:
Q = mcΔT, where ΔT represents the change in temperature.
In the case of the iron block:
m = 75 g
c = 0.449 J/g °C
ΔT = 1535 - 25 = 1510 °C
Therefore,
Q = 75 g x 0.449 J/g °C x 1510 °C
= 50,849.25 Joules
<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>
When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
Learn more about balancing equation here:
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Answer:Noble gases:
are highly reactive.
react only with other gases.
do not appear in the periodic table.
are not very reactive with other elements.
Explanation:Noble gases:
are highly reactive.
react only with other gases.
do not appear in the periodic table.
are not very reactive with other elements.
