When we convert the given mass in grams and volume in liters to m/v percent, we recall that m/v percent is expressed as grams/100 milliliters. In this case the expression becomes (50 grams/ 2500 L)*(0.1L/100ml), that is equal to 0.002 grams/ 100 mL. Hence the the concentration is equal to 0.2 m/v percent.
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
The molarity of Sr(OH)2 solution is = 0.1159 M
calculation
write the equation for reaction
that is, Sr(OH)2 +2HCl→ SrCl2 + 2 H2O
then finds the mole of HCl used
moles = molarity x volume
=40.03 x0.1159 = 4.639 moles
by use of mole ratio between Sr(OH)2 to HCL which is 1 :2 the moles of Sr(OH)2 is therefore = 4.639 x1/2 = 2.312 moles
molarity of Sr(OH)2 is = moles / volume
=2.312 /20 =0.1159 M
Answer:
It's False
Explanation:
In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.
Hope this helps you
<span>As mentioned, the isomerization of cyclopropane to propylene is a first-order process with a half-life of 19 min at 500°c. A first-order reaction kinetic rates means that the rate is constant throughout the reaction.
Thus, the time it takes for the partial pressure of cyclopropane to decrease from 1 atm to 0.125 atm at 500°c is </span><span>57 minutes.</span>
