Explanation:
A single-replacement reaction replaces one element for another in a compound.
A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.
A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.
Solubility rules are used to predict whether some double-replacement reactions will occur.
The final temperature of the mixture : 21.1° C
<h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Q in(gained) = Q out(lost)
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Q ethanol=Q water
mass ethanol=

mass water =

then the heat transfer :

Answer:
21 g/mL
Explanation:
To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.
It’s c because a wave cycle is repeating patterns all over again and again
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles