Answer:
B. Controlled variables
Explanation:
The Dependent sample t-test compares the mean score of measurements in one group to that of another other group. It mainly used when analyzing comparable sample units as it pairs repeatable observations within a time frame.
Controlled variables are factors that are not tested and remain constant. Some examples of controlled variables are temperature, pressure, volume etc
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
The volume of acetone, in milliliters, has a mass of 44.2 g is 55.96mL.
<h3>HOW TO CALCULATE VOLUME?</h3>
The volume of a substance can be calculated by dividing the mass by its density. That is;
Volume (mL) = mass (g) ÷ density (g/mL)
According to this question, acetone is a solvent with density of 0.7899g/mL and mass of 44.2g. The volume is calculated as follows:
Volume = 44.2g ÷ 0.7899g/mL
Volume = 55.96mL
Therefore, the volume of acetone, in milliliters, has a mass of 44.2g is 55.96mL.
Learn more about volume at: brainly.com/question/1578538
NH3 is a gas at room temperature. Solutions of "ammonia" are made by bubbling NH3 through water, so there's no "pure ammonia" unless you are talking about the gas.
Answer:
0.25 g of U-235 isotope will left .
Formula used :
where,
N = amount of U-235 left after n-half lives = ?
= Initial amount of the U-235 = 1.00 g
n = number of half lives passed = 2
0.25 g of U-235 isotope will left .
