Which of the following aqueous solutions

Use Table 1: Analysis of Decay in the laboratory guide to answer this question: Scientists find a piece of wood that is thought

sashaice [31]

1/2=5750 years, 1/2(1/2)=1/4, (1/2)(1/2)(1/2)=1/8, (1/2)(1/2)(1/2)(1/2)=1/6
4 halflives have passed so 4(5750)=23000 years since the tree was chopped down
1000000 atoms (1/2)=500000 atoms(1/2)=250000(1/2)=125000(1/2)=62500 atoms would remain in the wood after 4 halflives
Dinosaurs became extinct around 62 million years ago, so if 14C's half life has a value of 5750 years, it would be gone or in such small amounts that dating would be ineffective today.
As Potassium decays into Argon in 1.3 billion years, apart from volcanic activity, it would enable geologists to effectively date things that are really, really, really old.

8 0

3 years ago

Homeostasis is the balance between your internal and external environments. We

topjm [15]

Answer:

They are already matched for you. It goes

1.

2.

3.

in the order of the questions.

8 0

3 years ago

Which of the following can take the shape of the container it is in but has a fixed volume? (3 points)

IrinaK [193]

Easy, that's liquid sir.

4 0

3 years ago

Read 2 more answers

Determine the volume of occupied by 4 moles of carbon dioxide gas at

valentina_108 [34]

Answer:

Vol of 4 moles CO₂(g) at STP = 89.6 Liters

Explanation:

STP

P = 1 Atm

V =

T = 0°C = 273 K

n = 4 moles

R = 0.08206 L·Atm/mol·K

Using Ideal Gas Law PV = nRT => V = nRT/P

V = (4 moles)(0.08206 L·Atm/mol·K)(273 K)/(1 Atm) = 89.6 Liters

5 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

4 years ago