I think the answer would be d but not 100% sure
<span>The coyote (Canis latrans) and the gray wolf (Canis lupus) are the most similar species in the table.</span>
The pressure of the gas in the flask (in atm) when Δh = 5.89 cm is 1.04 atm
<h3>Data obtained from the question</h3>
The following data were obtained from the question:
- Atmospheric pressure (Pa) = 730.1 torr = 730.1 mmHg
- Change in height (Δh) = 5.89 cm
- Pressure due to Δh (PΔh) = 5.89 cmHg = 5.89 × 10 = 58.9 mmHg
- Pressure of gas (P) =?
<h3>How to determine the pressure of the gas</h3>
The pressure of the gas can be obtained as illustrated below:
P = Pa + PΔh
P = 730.1 + 58.9
P = 789 mmHg
Divide by 760 to express in atm
P = 789 / 760
P = 1.04 atm
Thus, the pressure of the gas when Δh = 5.89 cm is 1.04 atm
Learn more about pressure:
brainly.com/question/22523697
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Missing part of question:
See attached photo
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
