Answer: -
2.5 mL
Explanation: -
Volume of final solution = 10 mL
Strength of final solution = 10 mg / mL
Amount of drug required = Volume of final solution x strength of final solution
= 10 mL x 10 mg/ mL
= 100 mg.
Strength of supplied drug = 40mg/ mL
Volume of supplied drug required = 
= 2.5 mL
Thus to make 10 ml of a 10 mg/ml solution. 2.5 mL will be needed.
Answer:
The electrons of an atom are typically divided into two categories: valence and core electrons. Valence electrons occupy the outermost shell or highest energy level of an atom while core electrons are those occupying the innermost shell or lowest energy levels
Explanation:
i hope u get it :))
Answer:
Yep, this is my new pick up line
Explanation:
D = 0.2 g / ml = 0.2 g / cm³
For example, density of steel is 7.85 g / cm³.
Density of pure water is 1.0 g/cm³. An object which has a density < 1.0 g/cm³ will float in water.
Answer: Material that has a density of 0.2 g/ml ( 0.2 g/cm³ ) is good for making couch cushions.
Answer:
63.02 g.
Explanation:
- Na reacts with Cl₂ according to the balanced equation:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear that 2 mole of Na react with 1 mole of Cl₂ to produce 2 moles of NaCl.
- Firstly, we need to calculate the no. of moles of Na and Cl₂:
no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.
no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.
- From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).
<em>So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".</em>
The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.
∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.
- To get the no. of grams of produced NaCl:
<u><em>using cross multiplication:</em></u>
2 mol of Na produce → 2 mol of NaCl, from the stichiometry.
∴ 0.826 mol of Na produce → 0.826 mol of NaCl.
∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.
∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.