Fifty percent. plz brainliest my answer
So let's use some equations to represent the data [let R= cost of ring & B= cost of bracelet]
R= B + $ 36 .... (1)
B=

× R ... (2)
By using simultaneous equations to solve for B and R.
Substitute eq. (1) into eq. (2)
B =

× (B + $36)
B =

B +

⇒ B = $48
By substituting value of B into ea (1)
If R = B + $36
R = ($48) + $36
= $84
∴ <span> the total of the two items = R + B
= $84 + $48
</span> = $132
Answer:
1) The overlap of the p orbitals of the carbon-carbon π bond would be lost
Explanation:
Unlike simple bonds, a double bond can not rotate, since it is not possible to twist the ends of the molecule without breaking the π bond.
In the structure of but-2-ene present in the attachment, we can see the two isomers, <em>cis</em> and<em> trans</em>. These isomers cannot be interconverted by rotation around the carbon-carbon double bond without breaking the π bond.
Answer:
2KCl + F₂ → 2KF + Cl₂
Explanation:
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
2KCl + F₂ → 2KF + Cl₂
In this equation mass of reactant and product is equal. There are 2 potassium 2 chlorine and fluorine atoms on both side of equation it means mass remain conserved.
All other options are incorrect because mass is not conserved.
Mg₂ + LiBr ---> LiMg + Br
In this equation mass of magnesium is more on reactant side.
Na +O₂ ---> Na₂O
In this equation there is more oxygen and less sodium on reactant side while there is more sodium and less oxygen on product side.
H₂O ---> H₂ + O₂
In this equation there is less oxygen on reactant side while more oxygen on product side.
Here we have to calculate the amount of
ion present in the sample.
In the sample solution 0.122g of
ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + 
(Na)₂SO₄=2Na⁺ + 
Thus, BaCl₂+
= BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of
ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion (
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of
ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of
ion is present.