If the length of a string increases 2 times but the mass of the string remains constant, the original density will be multiplied by a factor mathematically given as
v=1.414 times
<h3>Will the new density of the string equal the original density multiplied by what factor?</h3>
Generally, the equation for the volume is mathematically given as
v = sqt(TL/m)
Where
density = mass/length
Therefore
In conclusion,v ratio will give us
v=1.414 times
Read more about volume
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Answer:
1.8 moles of O₂
Explanation:
The balance chemical equation for said double replacement (photosynthesis) reaction is as follow;
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
According to balance chemical equation,
6 moles of O₂ are produced by = 6 moles of CO₂
So,
1.8 moles of O₂ will be produced by = X moles of O₂
Solving for X,
X = 1.8 mol × 6 mol / 6 mol
X = 1.8 moles of O₂
Stoichiometric problems in which moles are given and moles or other reactant or product asked are the simplest problems. One should only write the balanced chemical equation and perform above method to find the required moles.
Answer:
C) sp2 and sp2
Explanation:
The hybridization depens on the ammount and type of bonds the atom analized has in the molecule.
For example:
- A C atom bonded to 4 H atoms has a sp3 hybridization.
- A C atom bonded to 2 H atoms and to 1 C with a double bond (like in ethene) has a sp2 hybridization
- A C bonded to 1 H and 1 C with a triple bond (like in ethyne) has a sp hybridization.
Analyzing the type and amount of unions of the nitrogen and the carbonyl you will be able to determine the hybridization.
In the imine, the N atom has a double bond to a C and a simple bond two other C, plus the lone pair of electrons (counts as a bond) so it will have a sp2 hybridization.
In the carbonyl, the C has two simple bonds to other C and a double bond to an oxygen atom. It will also have a sp2 hybridization
Answer:
2
Explanation:
Tried out 1, but couldn't get whole numbers on the right side then, so went up to 2, worked
all four numbrrs: 2, 7, 4, 6
The 7 for oxygen got adjusted in the last step of the thinking, because it's the simplest to adjust.
since there are different prime numbers (or rather numbers that don't share prime factors) in the set, the numbers can't be revived by the same number and still give whole numbers as results (they can't be smaller)
