Answer: option D. the ability of a base to react with a soluble metal salt.
Justification:
NaOH is a strong base, which means that in water it will dissociate according to this reaction:
- NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)
On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:
Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).
That is what the given reaction represents:
CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
↑ ↑ ↑ ↑
soluble metal salt strong base insoluble base solube salt
Answer:
D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.
Explanation:
Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.
<u>(1) Preparatory phase</u>
During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.
When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.
<u>(2) Pay-off phase</u>
During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.
It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.
Answer:
For the most part, non-metals (excluding Nobel gases) are the most likely to form covalent bonds. Pure covalent bonds are formed between atoms with the same electronegativity, ie. they are trying to hold on to the electrons in the bond with the same strength.
The molar mass of methylammonium bromide is 111u.
<h3>What is molar mass?</h3>
The molar mass is defined as the mass per unit amount of substance of a given chemical entity.
Multiply the atomic weight (from the periodic table) of each element by the number of atoms of that element present in the compound.
Add it all together and put units of grams/mole after the number.
Atomic weight of H is 1u
Atomic weight of N is 14u
Atomic weight of C is 12u
Atomic weight of Br is 79u
Calculating molar mass of 
=2(1 x3+ 14+12+ 1 x 3 +79) = 111u
Hence, the molar mass of methylammonium bromide is 111u.
Learn more about molar mass here:
brainly.com/question/12127540
#SPJ1
Answer:
The reaction must be spontaneous, the disorder of the system increases.
Explanation:
By the Second Law of Thermodynamics, a positive change in entropy is due to a net input heat, and entropy is a measure of the grade of disorder within the system. The net input heat means that resultant goes to the system from the surroundings.
By the First Law of Thermodynamics, a net input heat is due to a positive change in enthalpy.
The reaction is endothermic and spontaneous (since change in entropy is positive).
