substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>
Answer:
0.0000159
Explanation:
Divide 15.9 by 1000000, because 1 kilometer equals 1000000 millimeters.
Answer:
ΔHorxn = - 11.79 KJ
Explanation:
2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )
The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol
From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,
1 mol = 22.4l
x mol = 2.67l
Upon cross multiplication and solving for x;
x = 2.67 / 22.4 = 0.1192 mol
0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.
Amount of heat is given as;
ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)
Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.
ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)
ΔHorxn = - 47.17kj + 35.38kj
ΔHorxn = - 11.79 KJ
The answer is A. Gravity
The gravitioal force that an object exerted will increase as the object is increasing in sizre.
Which means that if object A has two-times more mass than object B, Object A will has twice as much gravitational pull if compared to object B.
Answer:The 1st and 2nd reactions are the example of oxidation -reduction.
Explanation:
Oxidation is basically when a species loses electrons and reduction is basically when the species gains electrons.
A reaction is known as an oxidation -reduction reaction only if oxidation and reduction simultaneously occur in the reaction. It basically means if a species is getting oxidized in the reaction then the other species present in the system must be reduced in the reaction.
Oxidation-reduction reactions are also known as redox reactions.
In the 1st reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of chlorine is 0 in reactants and in products is -1 so chlorine is reduced. Hence Na is oxidized and Cl is reduced so the reaction is a example of oxidation-reduction.
2Na(s)+Cl₂(g)→2NaCl(s)
In the second reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of Cu is +1 in reactant and 0 in products so Cu is reduced. Hence Na is oxidized and Cu is reduced so the reaction is an example of oxidation-reduction.
Na(s)+CuCl(aq)→NaCl(aq)+Cu(s)
In the third reaction the oxidation state of Na changes from +1 to +1 and that of Cu also changes from +1 to +1. So there is no change in oxidation state of the species present in reactants and products. Hence this reaction is not an example of oxidation and reduction.
