In alkene if two substituent and hydrogen are attached in the isomer may be cis or trans. When two or more substitution are attached to an alkene the isomer may be Z or E.All cis are Z isomer. The structure of (Z)-3-methy-3-heptene is as the following attachment
Answer: (3) 15
Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).
See figure 1
I hope it helps!
<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:
The answer to your question is: 0.373 g
Explanation:
Data
mass = ? g
atoms = 3 x 10 ²¹
AM = 74.92 g
Process
1 mol of As ------------------ 6.023 x 10²³ atoms
x ------------------ 3 x 10 ²¹ atoms
x = 4.98 x 10⁻³ moles
1 mol ------------------------ 74.92 g
4.98 x 10⁻³ moles------- x
x = (4.98 x 10⁻³ x 74.92)/1
x = 0.373 g of As
