Answer:
5.90x10²² atoms of gold are present in the cube
Explanation:
First, we must find the volume of the cube in cm³. With density we can find the mass of the gold and the moles using its molar mass. As 1 mol = 6.022x10²³ atoms we can find the number of atoms:
<em>Volume in cm³:</em>
(10.0mm)³ = 1000mm³
<em>1mm³ = 0.001cm³</em>
1000mm³ * (0.001cm³ / 1mm³) = 1cm³
<em>Mass gold:</em>
1cm³ * (19.3g/cm³) = 19.3g Gold
<em>Moles Gold:</em>
19.3g * (1mol / 197.0g) = 0.0980 moles Gold
<em>Atoms gold:</em>
0.0980 moles Gold * (6.022x10²³ atoms / mol) =
<h3>5.90x10²² atoms of gold are present in the cube</h3>
The atom that has the largest atomic radius is Fr.
Answer:
Tube 2: 8.26 * 10^-3; Tube 4: 6.83 * 10^-5
Explanation:
In the serial dilutions for MIC test, the volume of nutrient broth in each tube should be equal: 5.0 mL. And the volume of agent in each dilution should also be similar: 0.5 mL.
The serial dilutions was as following:
- Tube 1: 0.5/5.5
- Tube 2: 0.5 mL of tube 1 was diluted with 5.0 mL broth. Then, the dilution of tube 2 is (1:11) * (1:11) = (0.5/5.5) * (0.5/5.5) = 1:121 = 8.26 * 10^-3
- Tube 3: We perform the similar calculation. Thus, the result is 1:1331 = 7.51 * 10^-4
- Tube 4: It is 1:14641 = 6.83 * 10^-5.
ANSWER:
balanced equations are:
Mg + 2 HCl ----> MgCl2 +H2
and
2 NaCl ----> 2 Na + Cl2.
EXPLANATION : SEE PICS