Draw the mechanism arrows for both propagation steps for the radical addition of HBr to the alkene. When drawing single-headed r
adical arrows, this software requires that they meet at one atom (not in space between atoms like you may do in class). In the second box you will need to draw the first product and another reactant. In the last box you will need to draw an additional product.
1 answer:
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Answer:
The answer to the question is
The star’s approximate radial velocity is 68.52 km/s
Explanation:
To solve the
The formula is
where
= velocity of the star
λ = Star's spectrum wavelength = 656.45 nm
= Rest wavelength = 656.30 nm
c = Speed of light = 299 792 458 m / s
Therefore we have
or
= 68518.7699 m/s or 68.52 km/s
Hello!
We have the following data:
m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)
First we find the solute mass (m1), knowing that:
20% m/m = 20g/100mL
20 ------ 100 mL (0,1 L)
y g --------------- 1 L
y = 20/0,1
y = 200 g --> m1 = 200 g
Let's find Solute's Molar Mass, let's see:
M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol
We must find the volume of the solvent and therefore its mass (m2), let us see:
d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?
<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)
If: 1 mL = 1g
<span>Then the mass of the solvent is:
</span>
m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg
Now, we apply all the data found to the formula of Molality, let us see:
_________________________________
_________________________________
<span>Another way to find the answer:
</span>
We have the following data:
W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol
<span>Let's find the number of mols (n), let's see:
</span>
Now, we apply all the data found to the formula of Molality, let us see:
I hope this helps. =)
We have the following data:
m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)
First we find the solute mass (m1), knowing that:
20% m/m = 20g/100mL
20 ------ 100 mL (0,1 L)
y g --------------- 1 L
y = 20/0,1
y = 200 g --> m1 = 200 g
Let's find Solute's Molar Mass, let's see:
M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol
We must find the volume of the solvent and therefore its mass (m2), let us see:
d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?
<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)
If: 1 mL = 1g
<span>Then the mass of the solvent is:
</span>
m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg
Now, we apply all the data found to the formula of Molality, let us see:
_________________________________
_________________________________
<span>Another way to find the answer:
</span>
We have the following data:
W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol
<span>Let's find the number of mols (n), let's see:
</span>
Now, we apply all the data found to the formula of Molality, let us see:
I hope this helps. =)