Answer:
5.3 × 10⁻³ kg
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist adds 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>
<em />
We have 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution. The moles of zinc nitrate are:
0.1350 L × 0.21 mol/L = 2.8 × 10⁻² mol
The molar mass of zinc nitrate is 189.36 g/mol. The mass corresponding to 2.8 × 10⁻² moles is:
2.8 × 10⁻² mol × 189.36 g/mol = 5.3 g
1 kilogram is equal to 1000 grams. Then,
5.3 g × (1 kg/1000 g) = 5.3 × 10⁻³ kg
Answer:
magnesium chloride (no prefixes)
The correct answer is C because diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration.
HOPE THIS HELPS!
HAVE A GR8 DAY ;-)
Answer:
Infrared thermography
Explanation:
Infrared thermography is equipment or method, which detects infrared energy emitted from object, converts it to temperature, and displays image of temperature distribution. ... We call our equipment as infrared thermography considering such generalization of the terminology.
Answer:
Biggest Radii V²⁺ > V³⁺ > V⁴⁺ > V⁵⁺ Smallest Radii
General Formulas and Concepts:
- Periodic Trends: Atomic/Ionic Radii
- Coulomb's Law
Explanation:
The Periodic Trend for Atomic Radii is down and to the left. Therefore, the element with the largest radius would be in the bottom left corner of the Periodic Table.
Anions will always have a bigger radii than the parent radii. When we add e⁻ to the element, we are increasing the e⁻/e⁻ repulsions. This will cause e⁻ to repel themselves more and thus create more space, increasing the radii size.
Cations will always have smaller radii than the parent radii. When we remove e⁻ from the element, we are decreasing e⁻/e⁻ repulsions. Since there are less e⁻, there is no need for more space and thus decreases the radii size.
Since Cations are smaller than the parent radii, the more e⁻ we remove, the smaller it will become.
Therefore, the least removed e⁻ Vanadium would be the largest and the most removed e⁻ Vanadium would be the smallest.
