8.9 g/cm3 since density equals mass/volume or m/v you just divide 89 by 10 to get 8.9
Answer:
pressure of the oxygen gas is 7.88atm
Explanation:
Moles of oxygen in cylinder = 2.80 mol
volume=8.58L
temperature=294K
molar gas constant(R)= 0.0821 L atm / (mol K)
The pressure is calculated using ideal gas law:
PV = nRT
making Pressure subject of the formula
P
=
nRT
/V
=
2.80× 0.0821 ×294 /8.58 L
P = 7.88 atm
Answer:
The correct answer is -0.129 kJ
Explanation:
In the given case, the cooling of the balloon is done by withdrawing 0.784 J of heat, and the work done by the atmosphere on the balloon is 655 J. First, there is a need to transform kJ into J, 1 kJ = 1000 J. So, 0.784 kJ would be 784 J.
The ΔE or the change in the internal energy can be calculated by using the formula, ΔE = q + w ----- (1).
In the given case, q refers to the heat moved out of the system, that is, the value of q would be less than 0 or will be a negative quantity. Therefore, the heat moved out of the system will be -784 J. On the other hand, as the work or w is done on the system, therefore, the value of w would be more than 0 or will be a positive quantity. Thus, the value of w will be +655 J.
Now putting the values in the equation (1) we get,
ΔE = -784 J + 655 J
ΔE = -129 J or -0.129 kJ
As the change in internal energy comes out to be a negative value, therefore, the process is considered exothermic.
Answer:
7.75 Calories per gram of candy
Explanation:
The equation of a calorimeter is:
Q = C×ΔT
Where Q is heat involved in the reaction, C is heat capacity of the calorimeter (46.80kJ K⁻¹) and ΔT is change in temperature (2.91°C). Replacing:
Q = 46.80kJ K⁻¹ ₓ 2.91K
Q = 136.2 kJ are produced. If 4.184 kJ = 1kCal:
136.2 kJ ₓ (1kCal / 4.184 kJ) = <em>32.55 kCal per 4.20g of candy. </em>Thus, Calories / g of candy are:
32.55 kCal / 4.20g = <em>7.75 Calories per gram of candy</em>