Answer:
The new temperature will be 2546 K or 2273 °C
Explanation:
Step 1: Data given
The initial temperature = 1000 °C =1273 K
The volume = 20L
The volume increases to 40 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 20L
⇒with T1 = the initial temperature = 1273 K
⇒with V2 = the increased volume = 40L
⇒with T2 = the new temperature = TO BE DETERMINED
20L/ 1273 K = 40L / T2
T2 = 40L / (20L/1273K)
T2 = 2546 K
The new temperature will be 2546 K
This is 2546-273 = 2273 °C
Since the volume is doubled, the temperature is doubled as well
95 percent of species known are invertebrates
Have about 5 beakers all with different temperatures of water. Put in a teaspoon of salt at a time and when it stops dissolving stop adding and record how much salt it took. It should be more salt as the temperature rises. The independent variable is the waters temperature. The dependent variable is how much salt is used. Make sure that there is the same amount of water in each beaker. Or else it won’t work.
Concentration of Ni in 20mL = 5.28ppm x dilution factor = 5.28 x 100/5 = 105.6 ppm = 105.6 mg/L
molar mass of Ni = 58.6934 g
<span>Molarity of Ni = 100.40 x 10^{-3} / 58.6934 = 1.71 x 10^{-3} M = 1.71 mM. </span>
% mass Na = (g Na / g H2O) x 100
<span>g Na = (% mass Na)(g H2O) / 100 = (5.2 x 10^-4 g Na / g H2O) </span>
<span>2.4 g Na x (1 g H2O / 5.2 x 10^-4 g Na) = 4600 g Na </span>
<span>4600 g Na x (1 mL H2O / 1 g H2O) = 4600 mL H2O = 4.6 L H2O which is a little more than a gallon</span>
