Mercury-197 has a half life of 3 days. Starting with 300 grams, how much remains in 3 weeks? (Round to two places)

Answers

Answer 1

Answer: each > = 3 days
300g > 150g > 75g > 37.5g > 18.75g > 9.375g > 4.6875g > 2.34375g
Thus,mass remaining = 2.34g(2 d.p.)

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Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is sometimes difficult at first for people to understand.

Answers

The correct answer for the question that is being presented above is this one: "A rolling ball eventually slows down and comes to a stop." Newton’s first law of motion was a giant leap forward in scientific thought during Newton’s time. Even today, the idea is sometimes difficult at first for people to understand.

Mg(s) + 2HCI(aq) -> MgCl2(aq) + H2(g) a. How many grams of HCI are consumed by the reaction of 2.50 moles of magnesium? Given: Requested:

Answers

Answer:2.50 moles of magnesium will consume approximately 182.30 grams of hydrochloric acid in the given reaction.

Explanation:To find out how many grams of hydrochloric acid (HCl) are consumed when 2.50 moles of magnesium (Mg) react with it, you can use stoichiometry and the balanced chemical equation:

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

From the balanced equation, you can see that 1 mole of magnesium (Mg) reacts with 2 moles of hydrochloric acid (HCl).

Now, let's use this information to calculate the moles of HCl required to react with 2.50 moles of Mg:

Moles of HCl = (2.50 moles Mg) * (2 moles HCl / 1 mole Mg)

Moles of HCl = 2.50 moles * 2

Moles of HCl = 5.00 moles

Now that we know we need 5.00 moles of HCl, we can calculate the grams of HCl needed using the molar mass of HCl:

The molar mass of HCl is the sum of the atomic masses of hydrogen (H) and chlorine (Cl):

Molar mass of HCl = 1.01 g/mol (for hydrogen) + 35.45 g/mol (for chlorine)

Molar mass of HCl = 36.46 g/mol

Now, calculate the grams of HCl:

Grams of HCl = (5.00 moles) * (36.46 g/mol)

Grams of HCl = 182.30 grams

So, 2.50 moles of magnesium will consume approximately 182.30 grams of hydrochloric acid in the given reaction.

How many grams of O2(g) are needed to completely burn 73.7 g of C3H8(g)?

Answers

The balanced reaction is:

C3H8 + 5O2 = 3CO2 + 4H2O

We are given the amount of C3H8 to be reacted with O2. This will be the starting point of the calculations.

73.7 g C3H8 ( 1 mol C3H8/ 44.1 g C3H8 ) ( 5 mol O2 / 1 mol C3H8 ) ( 32.0 g O2 / 1 mol O2 ) = 267.39 g O2

Answer: 267.2 grams

Explanation:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of propane}=(73.7g)/(44.1g/mol)=1.67moles$

According to stoichiometry:

1 mole of $C_3H_8$ require 5 moles of $O_2$

Thus 1.67 moles of $C_3H_8$ will require = $(5)/(1)* 1.67=8.35moles$ of $O_2$

Mass of $O_2=moles* {\text {Molar mass}}=8.35moles* 32g/mol=267.2$

Thus 267.2 grams of oxygen are required.

A solute is a mixture in which particles can be seen and easily separated. Please select the best answer from the choices provided T F

Answers

F because solute is the one which is mixed to the other for a solution and after mixing I can never been seen neither separated !!

Rainforests are typically responsible for global oxygen turnover

Answers

false...........................................

Answer:

false

Explanation:

A laboratory requires 2.0 L of a 1.5 M solution of hydrochloric acid (HCl), but the only available HCl is a 12.0 M stock solution. How could you prepare the solution needed for the lab experiment? show all work to find your answer

Answers

Answer: The volume of stock HCl solution required to make laboratory solution will be 0.25L

Explanation:

To calculate the volume of stock solution required to make the laboratory solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock solution.

$M_2\text{ and }V_2$ are the molarity and volume of laboratory solution.

We are given:

$M_1=12M\nV_1=?L\nM_2=1.5M\nV_2=2L$

Putting values in above equation, we get:

$12* V_1=1.5* 2\n\nV_1=0.25L$

Hence, the volume of stock HCl solution required to make laboratory solution will be 0.25L

The amount of the solute is constant during dilution. So the mole number of HCl is 2*1.5=3 mole. The volume of HCl stock is 3/12=0.25 L. So using 0.25 L stock solution and dilute to 2.0 L.

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