Answer:
1.0 *10^(-4) mol
Explanation:
For gases:
n1/n2 = V1/V2
n1/3.8*10^(-4) mol = 230 mL/ 860 mL
n1 = 3.8*10^(-4)*230/860 = 1.0 *10^(-4) mol
The question is incomplete, the complete question is shown in the image attached
Answer:
A and B
Explanation:
The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.
The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.
Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.
In mammals and amphibians? An enucleated egg, a donor nucleus (preferably from an early developmental stage such as a blastocyst), and a means to stimulate the egg to be activated as if it had just been fertilized (poking with a needle is sometimes enough)
Or cloning into a vector as in at the level of DNA?
A vector with positve and negative selection markers (like antibiotic resistance and drug susceptibility), an insert, DNA ligase and restriction enzymes, buffer for restriction and ligation. Or if you are doing cloning by PCR, then you need primers, template, nucleotides, RNA pol like Taq polymerase etc.
Answer:
We need 10.14 grams of sodium bromide to make a 0.730 M solution
Explanation:
Step 1: Data given
Molarity of the sodium bromide (NaBr) = 0.730 M
Volume of the sodium bromide solution = 135 mL = 0.135 L
Molar mass sodium bromide (NaBr) = 102.89 g/mol
Step 2: Calculate moles NaBr
Moles NaBr = Molarity NaBr * volume NaBr
Moles NaBr = 0.730 M * 0.135 L
Moles NaBr = 0.09855 moles
Step 3: Calculate mass of NaBr
Mass NaBr = 0.09855 moles * 102.89 g/mol
Mass NaBr = 10.14 grams
We need 10.14 grams of sodium bromide to make a 0.730 M solution
According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal.
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