Decide whether a chemical reaction happens in either of the following situations. If a reaction does happen, write the chemical
1 answer:
Answer:
Explanation:
Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.
A more reactive element is one which can easily lose or gain electrons as compared to the other element.
As Magnesium is more reactive than tin, it can easily displace tin from its salt solution and form magnesium chloride and tin in elemental form.
The balanced chemical equation is:
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Answer: The final temperature of both the weight and the water at thermal equilibrium is .
Explanation:
The given data is as follows.
mass = 7.62 g,
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q =
=
Now, heat gained by lead will be calculated as follows.
q =
=
According to the given situation,
Heat lost = Heat gained
=
T =
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is .
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. =)