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/Explanation:
Dimitri Ivanovich Mendeleev
Please mark me BRAINLIEST! please!
Explanation:
Work done = force * perpendicular distance
= 1 * 0.01 = 0.01 joules
You were supposed to attach a picture so I know what the question id
