Answer: 1.48 atmosphere
Explanation:
Pressure in kilopascal = 150
Pressure in atmosphere = ?
Recall that 1 atmosphere = 101.325 kilopascal
Hence, 1 atm = 101.325 kPa
Z atm = 150 kPa
To get the value of Z, cross multiply
150 kPa x 1 atm = 101.325 kPa x Z
150 kPa•atm = 101.325 kPa•Z
Divide both sides by 101.325 kPa
150 kPa•atm/101.325 kPa = 101.325 kPa•Z/101.325 kPa
1.48 atm = Z
Thus, 150 kPa is equivalent to 1.48 atmospheres
Answer: b. The waste generated is hazardous and must be disposed of.
c. Nuclear material can be spilled into the ocean if reactors are near the coast.
d. A large amount of cold water is generated, which must be stored somewhere.
Explanation:
The main environmental costs for produced during the nuclear power plant consists of procurement of fuel and the thermal load is also produced with cold water discharge in the sea. This can contaminate the sea, hence, must be stored somewhere. The nuclear waste consists of radioactive substances which are hazardous for the environment. The nuclear based electricity does not produce carbon dioxide.
Answer:In chemistry, a valence electron is an outer shell electron that is associated with an atom, and that can participate in the formation of a chemical bond if the outer shell is not closed; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
Explanation: hope this helped
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>