Answer : The specific heat of metal is
.
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of metal = ?
= specific heat of water = 
= mass of metal = 129.00 g
= mass of water = 45.00 g
= final temperature = 
= initial temperature of metal = 
= initial temperature of water = 
Now put all the given values in the above formula, we get


Therefore, the specific heat of metal is
.
Answer:
a
The expected value is = 39 units
The standard deviation is = 0.1212 unites
b
The probability is = 0.2047 units
Explanation:
The explanation of this answer is shown on the first uploaded image
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
- The independent variable (IV) is the lemon juice mixture
- The dependent variable (DV) is the appearance of the green slime on the shower
- The control variable (CV) are time taken to spray, the amount of spray
- The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
- The control group (CG) is the side of the shower sprayed with water.
INDEPENDENT VARIABLE
- Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.
DEPENDENT VARIABLE:
- Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.
CONTROL VARIABLE:
- Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray
CONTROL GROUP
- Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.
EXPERIMENTAL GROUP:
- Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.
Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:
- The independent variable (IV) is the lemon juice mixture
- The dependent variable (DV) is the appearance of the green slime on the shower
- The control variable (CV) are time taken to spray, the amount of spray
- The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
- The control group (CG) is the side of the shower sprayed with water.
Learn more: brainly.com/question/17498238?referrer=searchResults
Answer:
Kp = 0.022
Explanation:
<em>Full question: ...With 2.3 atm of ammonia gas at 32. °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 0.69 atm. </em>
<em />
The equilibrium of ammonia occurs as follows:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where Kp is defined as:

<em>Where P represents partial pressure of each gas.</em>
<em />
As initial pressure of ammonia is 2.3atm, its equilibrium concentration will be:
P(NH₃) = 2.3atm - 2X
<em>Where X represents reaction coordinate</em>
<em />
Thus, pressure of hydrogen and nitrogen is:
P(N₂) = X
P(H₂) = 3X.
As partial pressure of hydrogen is 0.69atm:
3X = 0.69
X = 0.23atm:
P(NH₃) = 2.3atm - 2(0.23atm) = 1.84atm
P(N₂) = 0.23atm
P(H₂) = 0.69atm

<h3>Kp = 0.022</h3>