Answer:
use coefficients and subscripts to determine how many atoms are in a compound. If there is no subscript or coefficient, assume it is 1. If there is a coefficient, multiply it with the subscripts. For counting cations and anions, determine first which is the anion and cation (anion = nonmetal, cation = metal), then count the number of that ion.
Example:
NaCl
one atom of Na, one atom of Cl. Since Na is a metal, it is a cation. Cl is a nonmetal, so it is an anion.
2CaCl2
2 atoms of Ca, 4 atoms of Cl. There are 2 cations, since Na is a metal, and 4 anions since Cl is a nonmetal
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 .
To learn more about variables here
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Answer:
The ability to do work is energy.
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Answer:
50,849.25 Joules
Explanation:
The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:
Q = mcΔT, where ΔT represents the change in temperature.
In the case of the iron block:
m = 75 g
c = 0.449 J/g °C
ΔT = 1535 - 25 = 1510 °C
Therefore,
Q = 75 g x 0.449 J/g °C x 1510 °C
= 50,849.25 Joules
<em>Hence, </em><em>50,849.25 Joules </em><em> of heat must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C</em>
