I think the correct answer from the choices listed above is option A. The structural level of a protein least affected by a disruption in hydrogen bonding is the primary level. The other levels are very much affected by hydrogen bonding. Hope this answers the question.
Answer:
C
Explanation:
The answer is C because only that amount can move
Answer:
S/.486 es el valor del anillo
Explanation:
Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.
Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:
90g × 59.1% = 53.19g Oro en el anillo
Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:
53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.
Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:
0.27mol × (S/.1800 / 1mol oro) =
<h3>S/.486 es el valor del anillo</h3>
Answer:
- <u>Cadmium has larger atomic radius than sulfur.</u>
Explanation:
Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.
Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:
- Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.
- Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.
Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:
- Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.
- Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.
So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:
- O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.
Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.
first we have to find the empirical formula of the compound
empirical formula is the simplest ratio of whole numbers of components making up a compound
for 100 g of the compound
N O
mass 46.7 g 53.3 g
number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol
moles = 3.34 mol = 3.33 mol
divide by the least number of moles
3.34/3.33 = 1.00 3.33/ 3,33 = 1.00
therefore number of atoms are
N - 1
O - 1
empirical formula is - NO
mass of empirical unit - 14 g/mol + 16 g/mol = 30 g
molecular formula is actual composition of elements in the compound
molecular mass - 60.01 g/mol
number of empirical units = molecular mass / empirical unit mass
= 60.01 g/mol / 30 g = 2
there are 2 empirical units
2(NO)
molecular formula = N₂O₂
