A(n) DRUG is a chemical substance used to treat, prevent, or diagnose a disease or alter one or more functions of the body.
Example: Tylenol (Acetaminophen), Is a Drug that is used to relieve pain, head aches, and reduce fever.
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Answer:
A(n) is a chemical substance used to treat, prevent, or diagnose a disease or alter one or more functions of the body.
Explanation:
Answer:
Water becomes hot when the chemical potential of the water is equal to the chemical potential of the water at the same temperature as the pan.
Answer : The amount left after 20 minutes is, 0.592 grams.
Explanation :
Half-life of Bromine-85 = 3 min
First we have to calculate the rate constant, we use the formula :
Now we have to calculate the amount left after decay.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time taken by sample = 20 min
a = initial amount of the reactant = 60 g
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the amount left after 20 minutes is, 0.592 grams.
B. Atomic nuclei possess shells similar to the electron shells of atoms.
C. Some atoms contained more protons than electrons.
The molar mass of NaBr is 102.9 grams and that of Pb(NO3)2 is 331.21 grams.
Answer: 193.4 grams
Explanation: According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number of particles.
To calculate the moles, we use the equation:
According to stoichiometry
1 mole of reacts with 2 moles of
0.94 moles of will react with= of
Mass of
Answer:
193.24g
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
Pb(NO3)2(aq) + 2NaBr(g) -> PbBr2(s) + 2NaNO3(aq)
Step 2:
Determination of the masses of Pb(NO3)2 and NaBr that reacted from the balanced equation. This is illustrated below:
Molar Mass of Pb(NO3)2 = 331.21g/mol
Molar Mass of NaBr = 102.9g/mol
Mass of NaBr from the balanced equation = 2 x 102.9 = 205.8g.
From the balanced equation,
Mass of Pb(NO3)2 that reacted = 331.21g
Mass of NaBr that reacted = 205.8g
Step 3:
Determination of the mass of NaBr that reacted with 311g of Pb(NO3)2. This is illustrated below:
From the balanced equation above,
331.21g of Pb(NO3)2 reacted with 205.8g of NaBr.
Therefore, 311g of Pb(NO3)2 will react with = (311x205.8)/331.21 = 193.24g of NaBr.
From the calculations made above, 193.24g of NaBr will react with 311g of Pb(NO3)2