Answer: A = 1560
B = 1.6
Explanation: brainlest please
<span>(a) what is the average volume (in cubic meters) required for each iron atom
For this case, the density of Iron would be </span>7.87g/cm³
<span>
V = 9.27 x 10^-26 kg / </span>7.87g/cm<span>³ ( 1 kg / 1000 g)
</span>V = 1.18 x 10-23 cm³<span>
(b) what is the distance (in meters) between the centers of adjacent atoms?
We assume the atoms as cube, so we use the volume of the cube to calculate the distance of the atoms.
V = </span>1.18 x 10-23 cm<span>³ = s</span>³
s = 2.28 x 10^-8 cm
Answer:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles]
0.03 = log [A-] / [0.0500moles]
1.017152 = [A-] / [0.0500moles]
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) =
<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
In general chemistry, isotopes are substances that belong to one specific element. So, they all have the same atomic numbers. But they only differ in the mass numbers, or the number of protons and neutrons in the nucleus. In a nutshell, they only differ in the number of neutrons.
For Nickel, there are 5 naturally occurring isotopes. Their identities, masses and relative abundance are listed below
Isotope Abundance Atomic Mass
Ni-58 68.0769% <span>57.9353 amu
Ni-60 </span>26.2231% <span>59.9308 amu
Ni-61 </span>1.1399 % <span>60.9311 amu
Ni-62 </span>3.6345% <span>61.9283 amu
Ni-64 </span>0.9256% <span>63.9280 amu
To determine the average atomic mass of Nickel, the equation would be:
Average atomic mass = </span>∑Abundance×Atomic Mass
Using the equation, the answer would be:
Average atomic mass = 57.9353(68.0769%) + 59.9308(26.2231%) + 60.9311(1.1399%) + 61.9283(3.6345%) + 63.9280(0.9256%)
Average atomic mass = 58.6933 amu
