Grady is doing an experiment about the solubility of sugar. He puts 100 milliliters of water in each of three beakers. He leaves the first beaker at room temperature, heats the second beaker to 60°C, and heats the third beaker until the water boils at 100°C. The variable Grady change on purpose in the experiment is the temperature of water in each beaker .
Variables in the experiment is the any factor that can exist in different types or amount. There are three types of variables: independent variable , dependent variable , controlled variable. The independent variable is the variable you changed in the experiment. dependent variable is that changes because of independent variable. the controlled variable is the constant one.
Thus, Grady is doing an experiment about the solubility of sugar. He puts 100 milliliters of water in each of three beakers. He leaves the first beaker at room temperature, heats the second beaker to 60°C, and heats the third beaker until the water boils at 100°C. The variable Grady change on purpose in the experiment is the temperature of water in each beaker .
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Hydrogen and carbon atoms
Explanation:
Hydrocarbons are molecules that are made up of hydrogen and carbon atoms only.
They are made up of covalent bonds between the carbon and hydrogen atoms.
There is a wide range of hydrocarbons due to the unique properties of carbon.
Most of the hydrocarbons comes from crude oil, natural gas and coal.
When hydrocarbons burn, they produce water and carbon dioxide alone.
They are used as fuel in industrial processes because they produce heat energy.
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Answer:
At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.
Explanation:
Given data:
At volume = ?
Mass of NaOH = 12.5 g
Molarity of solution = 0.540 M
Solution:
First of all we will calculate the number of moles of sodium hydroxide.
Number of moles = mass/molar mass
Number of moles = 12.5 g / 40 g/mol
Number of moles = 0.3125 mol
Volume of NaOH:
Molarity = number of moles / volume in L
Now we will put the values.
0.540 M = 0.3125 mol / volume in L
volume in L = 0.3125 mol / 0.540 mol/L
volume in L = 0.58 L
