A possible cause of a large percentage of error in an
experiment where MgO is produced from the combustion of magnesium would be not all of the Mg has
completely reacted. <span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
Tornado is a spiral of debris
Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2
Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L
Answer:
Heat required = 13,325 calories or 55.75 KJ.
Explanation:
To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water
Which equals ,
Q = mL,
Where, m is the mass of water present
L = specific latent heat of vaporization
Here , m= 25 gram
L equals to 533 calories (or 2230 Joules)
So, Q = 25×533 = 13,325 Calories
Or , Q = 55,750 Joules = 55.75 KJ
so, Heat required = 13,325 calories or 55.75 KJ.
