Answer:
The correct option is: stable environment
Explanation:
According to the Darwin's theory, natural selection is the concept by which all the small useful variations of traits are preserved.
According to Darwin, there are three <u>necessary and sufficient conditions</u> for the occurrence of natural selection:
1. struggle for existence
2. variation
3. inheritance
These conditions are said to be necessary because if these conditions are not satisfied then natural selection does not occur.
These conditions are said to be sufficient because if these conditions are satisfied, then natural selection will most definitely occur.
First one is balanced because all of the elements in reactant and product have equal number of atoms.
second one is not balanced therefore you need to add coefficient to balance it. so its CaCl2 + 2NaHCO3 = CO2 + CaCO3 + H2O + 2NaCl
both of them seem double replacement reaction
Hope it helps!
~ Rabin
Answer:
It depends on the number of significant figures you are changing to
Explanation:
6.02×1023=6158.46
1 sig fig = 6000
2 sig fig = 6200
3 sig fig = 6160
4 sig fig = 6158
5 sig fig = 6158.5
6 sig fig = 6158.46
When solving significant figures you have to consider the number after each number like in the case of changing to two sig fig the number following one is five and when the number is up to or greater than five you add a value of one to the number before it. But in a case where the number is less than five you just leave it like that like in the case of changing to one sig fig
Answer:
yes, a compound can have more than one carbon-carbon double bond
Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
- A weak acid and its conjugate base.
- A weak base and its conjugate acid.
Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
