The answer is: D.unstable nuclei emitting high-energy particles as they formed more stable compositions.
Those high-energy particles are alpha particles
, beta particles
, gamma radiation.
For example, the decay chain of ²³⁸U is called the uranium series.
Decay start with U-238 and ends with Pb-206. There are several alpha and beta minus decays.
Antoine Henri Becquerel (1852 – 1908) was a French physicist and the first person to discover evidence of radioactivity.
Becquerel wrapped fluorescing crystal (uranium salt potassium uranyl sulfate) in a cloth, along with the photographic plate and a copper Maltese cross.
Several days later, he discovered that a image of the cross appeared on the plate.
The uranium salt was emitting radiation.
Because of this discovery, Becquerel won a Nobel Prize for Physics in 1903, which he shared with Marie Curie and Pierre Curie.
I believe the answer is C) there is an obvious reasoning for this all you have to do is eliminate answers that don't seem right for example, A)the plates are all moving the same direction every plate moves in different directions. B) The plates are all the same size. Well, it's really obvious that that is not true because every plate has its different shape and size. D) where two plates meet, they always move apart. If this were true, then we would never have earthquakes when plates meet earthquakes happen. so there for the answer is C)
A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is (
).
Utilize the titration method of in view that we're given the concentrations of every compound and the quantity of 
. Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.
- M1 = initial mass
- V1= initial volume
- M2 = final mass
- V2= final volume
- (0.138)(V1) = (0.205)x(26.0)
- V2=(0.205)x(26.0)\ 0.138
- V2 = 47.10 M/L
- The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L
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Answer:
e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi
a. Pyrophosphatase converts PPi and water into two Pi
b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP
Explanation:
Glycogen synthesis or glycogenesis is the process of synthesis of glycogen molecules from glucose molecules in living organisms. Glycogen is a polysaccharide storage form of glucose and helps to store excess glucose in the body form use when required by the body.
The synthesis of glycogen involves sugar nucleotides. Sugar nucleotides are compounds in which a sugar molecule is attached to a nucleotide through phosphate ester bond, resulting in the activation of the sugar molecule. The sugar nucleotides then are used as substrates for the polymerization of the monosaccharide sugars into disaccharides, oligosaccharides and polysaccharides.
In the synthesis of glycogen, glucose-6-phosphate from phosphorylation of free glucose by hexokinase is first isomerized to glucose-1-phosphate by phosphoglucomutase.
Glucose-1-phosphate is then converted to UDP-glucose by its reaction with UTP catalyse by UDP-glucose pyrophosphorylase. The reaction is favoured by the rapid hydrolysis of PPi produced to two molecules of inorganic phosphate by the enzyme pyrophosphatase.
Glycogen synthase then adds a glucose unit from UDP-glucose to a growing chain of glycogen, producing a larger glycogen molecule and free UDP.
Answer:
The coefficients are 6, 1, 3
Explanation:
HNCO →C3N3(NH2)3 + CO2
From the above equation, there are a total of 6 atoms of nitrogen on the right side and 1atom on the left. It can be balance by putting 6 in front of HNCO as shown below:
6HNCO → C3N3(NH2)3 + CO2
Now there are 6 atoms of carbon on the left side and 4 atoms on the right side. It can be balance by putting 3 in front of CO2 as shown below:
6HNCO → C3N3(NH2)3 + 3CO2
Now the equation is balanced as the numbers of atoms of the different elements on both sides of the equation are the same.
The coefficients are 6, 1, 3
