Base pairs of DNA are A with T, and C with G
here's how to remember:
Apple - Tree
Car - Garage
The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
Answer:
Your question is complex, because I think you wrote it wrong.
Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.
This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.
Simple junctions talks about compound organisms called ALKANS.
The double unions, in organic these compounds are called as ALQUENOS.
And as for the tertiary unions, the organic chemistry names them as ALQUINOS.
These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.
Explanation:
In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties
The answer would be b. Temperature of the solution increases
Temperature determines the kinetic energy of the water molecule. Higher temperature will cause the molecule to moves faster and the compound (KNO3) could break solute molecule easier make it become more soluble. A higher pressure will increase the solubility of a gas, not solid
Answer:
4.7 kJ/kmol-K
Explanation:
Using the Debye model the specific heat capacity in kJ/kmol-K
c = 12π⁴Nk(T/θ)³/5
where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K
Substituting these values into c we have
c = 12π⁴Nk(T/θ)³/5
= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5
= 9710.83(298 K/2219 K)³/5
= 1942.17(0.1343)³
= 4.704 J/mol-K
= 4.704 × 10⁻³ kJ/10⁻³ kmol-K
= 4.704 kJ/kmol-K
≅ 4.7 kJ/kmol-K
So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K