Answer:
D
Explanation:
because the higher the wavelength the shorter the pitch and the shorter the wavelength the higher the pitch
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
Answer:
Helium is in group 18 of the periodic table. How is helium different from the other elements in this group? = Helium atoms have 2 valence electrons, while atoms of the other elements in the group all have 8 valence electrons.
An emission spectrum will occur when = An electron releases energy and falls back to a lower energy level.
Explanation:
Answer:
0.1 g/dl
Explanation:
The standard curve is a graph that relates the absorbance at 400 nm with the concentration of haemoglobin in mg/dl. To obtain the concentration from the absorbance value, we enter in the x-axis (absorbance at 400 nm) with the value 0.40 (the line between 0.2 and 0.6), we extrapolate the line to the curve and read the correspondent value on y-axis (concentration in mg/dl): 100 mg/dl.
So, we convert the concentration from mg/dl to g/dl by dividing into 1000:
100 mg/dl x 1 g/1000 mg = 0.1 g/dl
Therefore, the concentration of haemoglobin of the patient is 0.1 g/dl.
Answer:
There are two major ways that molecules can be moved across a membrane, and the distinction has to do with whether or not cell energy is used. Passive mechanisms like diffusion use no energy, while active transport requires energy to get done.
Explanation: