greenhouse gases can cause Earth's atmosphere to trap more and more heat. This causes Earth to warm up.
The required mass of calcium bromide is 35.98 grams.
<h3>What is molarity?</h3>
Molarity is any solution is define as the number of moles of solute present in per liter of solution as;
M = n/V, where
- M = molarity = 4M
- V = volume = 45mL = 0.045L
Moles will be calculated by using the above equation as:
n = (4)(0.045) = 0.18 mole
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g
Hence required mass of CaBr₂ is 35.98 grams.
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Answer:
3 bonds are needed.
Explanation:
The electrons that are involved in chemical bonding are those in the outer shell of the highest energy level of the atom. The electron configuration of nitrogen (N) is 1s²2s²2p³. That means thy at each nitrogen atom has 5 valence electrons: 2 electrons in the 2s orbital and 3 electrons in the 2p orbital. To fullfil the octet, each nitrogen atom needs 3 electrons. So, they can share each other 3 electrons to form 3 simple bonds. Therefore, the nitrogen molecule (N₂) has 3 bonds involving 6 bonding electrons or a triple bond.
Answer:
a) The concentration of drug in the bottle is 9.8 mg/ml
b) 0.15 ml drug solution + 1.85 ml saline.
c) 4.9 × 10⁻⁵ mol/l
Explanation:
Hi there!
a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml
b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:
0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.
The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)
If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.
To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline
c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.
Let´s convert it to molarity:
0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l
