Answer:
0.15 l of 4.0 m stock KCl solution should betaken
Explanation:
N1V1=N2V2
6*0.1=V2*4
V2=0.15L
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
It means <span>a substance formed by chemical union of 2 or more elements or ingredients in definite proportion by weight.</span>
Answer:
The new pressure is 0.5 atm
Explanation:
Step 1: Data given
Volume of oxygen = 300 mL = 0.300 L
Pressure = 1.00 atm
Temperature = 300 K
The volume increases to 1000mL = 1.00 L
The temperature increases to 500 K
Step 2: Calculate the new pressure
(P1*V1)/T1 = (P2*V2)/T2
⇒with P1 = the initial pressure = 1.00 atm
⇒with V1 = the initial volume = 0.300 L
⇒with T1 = the initial temperature = 300 K
⇒with P2 = the new pressure = TO BE DETERMINED
⇒with V2 = the increased volume = 1.00 L
⇒with T2 = the increased temperature = 500 K
(1.00 atm* 0.300 L)/300 K = (P2 * 1.00L) / 500 K
P2 = (1.00 *0.300 * 500) / (300 *1.00)
P2 = 0.5 atm
The new pressure is 0.5 atm
