All categories

3 years ago

Carbon dating requires that the object being tested contain

Chemistry

1 answer:

luda_lava [24]3 years ago

4 0

Answer:

Organic Material

Explanation:

Carbon Dating is the process in which the age of a piece of organic matter is determined by the proportions of carbon isotopes it contains.

You might be interested in

Convert from 1.56×1030 particles of sodium chloride (NaCl) to grams of sodium chloride.

Anika [276]

Answer:

15.14×10⁷ g

Explanation:

Given data:

Number of particles of NaCl = 1.56×10³⁰ particles

Mass of sodium chloride = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ particles

1.56×10³⁰ particles × 1 mol / 6.022 × 10²³ particles

0.259 ×10⁷ mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.259 ×10⁷ mol × 58.44 g/mol

Mass = 15.14×10⁷ g

8 0

3 years ago

2 Na + Cl2 → 2 NaCl

Ksivusya [100]

Approximately 71.77 grams. I think so

8 0

3 years ago

Read 2 more answers

Which of the following statements supports the one gene-one enzyme hypothesis?A) A mutation in a single gene can result in a def

iragen [17]

Answer: its A

Explanation:

4 0

3 years ago

Read 2 more answers

Please help me with this question please!!!

Deffense [45]

Answer:

Explanation:

I looked up aromatic hydrocarbon and this one looks like a replica of benzene

3 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