Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
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Answer:
the physical and chemical properties of the products are different from the reactants
Explanation:
A chemical reaction involves the chemical combination of two or more elements/compounds called Reactants to give one or more different elements/compounds called Products. A chemical reaction occurs in such a way that the atoms of the reactants are restructured to form product(s) that is/are entirely different from the reactants.
In a chemical reaction, the physical and chemical properties of the products differ from that of the reactants since different chemical compounds/elements are formed as products. The physical properties of a substance, which include colour, melting and boiling point etc. will differ in the reactants and products formed. Also, the chemical structure and identity of the reactants will be changed to give rise to a different chemical property in the products.
Answer:
-290KJ/mol
Explanation:
ΔHrxn = ΔHproduct - ΔHreactant
ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}
ΔHrxn = 4(-1279) - [6(-286) - 3110]
= -5116 -(-1716-3110)
= -5116-(-4826)
= -5116 + 4826 = -290KJ/mol
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
