Answer:
67.91 g of CuCl2; 32.09 g of Cu.
Explanation:
The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.
The balanced chemical
reaction will be:
C4H8 + 6 O2 --> 4 CO2 + 4 H2O
We are given the amount of butene being combusted. This will be our
starting point.
136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2
To start this test, you need to identify the variables it presents. As you may already know, there are independent and dependent variables. Independent variables are those that act on a factor, influencing it to generate a result. In the case of this experiment, the independent variable is the completion of the homework. The dependent variable, in turn, is the factor that receives the influence of the independent variable, in this experiment this variable is the final grade you received in the course.
After that you must select a number of students, give them their homework and ask each student to complete a percentage of that amount. An example of this could be that you select 11 students and ask the first to complete 0% of the homework, the second student must complete 10%, the third 20% and so on, and the 11th student must complete 100% of the homework.
after that, note what was the final grade that each student received in the course and make a graph to show the results.
The y-axis of the graph must represent the dependent variable, while the x-axis must represent the independent variable. This way you will show the exact relationship between completing homework and the final grade of the course.
Answer:
They are in constant motion.
Explanation:
More energy\heat= more kinetic energy=more motion\movement
Answer:
187.34 atm
Explanation:
From the question,
PV = nRT.................. Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
make P the subject of the equation
P = nRT/V.............. Equation 2
n = mass(m)/molar mass(m')
n = m/m'............... Equation 3
Substitute equation 3 into equation 2
P = (m/m')RT/V............ Equation 4
Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L
Constant: m' = 2 g/mol, R = 0.082 atmL/K.mol
Substitute these values into equation 4
P = (46/2)(0.082×298)/3
P = (23×0.082×298)/3
P = 187.34 atm
