Among the following scenarios, which best demonstrates resilience when confronted with stress?

Thallium consists of 29.5%Th-203 and 70.5%Th-205. What is the relative atomic mass of thallium?

grin007 [14]

0.295 * 203 = 53.885
0.705 * 205 = 144.525
53.885 + 144.525 = 204.41
The relative atomic mass of Thallium is 204.41

7 0

3 years ago

It is desired to produce 2.25 grams of dichloromethane (CH2Cl2) by the following reaction. If the percent yield of dichlorometha

Allushta [10]

Answer:- 3.12 g carbon tetrachloride are needed.

Solution:- The balanced equation is:

$CH_4+CCl_4\rightarrow 2CH_2Cl_2$

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.

percent yield formula is:

percent yield = $(\frac{actual}{theoretical})100$

$65.5=(\frac{2.25}{theoretical})100$

$theoretical=(\frac{2.25(100)}{65.5})$

theoretical = 3.44 g

From balanced equation, there is 2:1 mol ratio between dichloethane and carbon tetrachloride.

Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.

$3.44gCH_2Cl_2(\frac{1molCH_2Cl_2}{84.93gCH_2Cl_2})(\frac{1molCCl_4}{2molCH_2Cl_2})(\frac{153.82gCCl_4}{1molCCl_4})$

= $3.12gCCl_4$

So, 3.12 grams of carbon tetrachloride are needed to be reacted.

8 0

3 years ago

19. I accidentally mixed bleach and ammonia when cleaning one day. It released 55 liters of deadly chlorine (CI) gas. How many g

user100 [1]

55,000 grams
Hope this helps you out :)

3 0

2 years ago

Can you pls answer this question cuz i don't know what is the answer on this..

Marysya12 [62]

Answer:

here's is ur answer in attachment

Explanation:

mark it brainliest if it helps you ❤️

3 0

3 years ago

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 13. g of butane is m

12345 [234]

Answer: The maximum amount of water that could be produced by the chemical reaction is 20.16 grams

Explanation:

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

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

For butane:

Given mass of butane = 13 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of butane}=\frac{13g}{58.12g/mol}=0.224mol$

For oxygen gas:

Given mass of oxygen gas = 70.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{70.9g}{32g/mol}=2.216mol$

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

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

By Stoichiometry of the reaction:

2 moles of butane reacts with 13 moles of oxygen gas

So, 0.224 moles of butane will react with = $\frac{13}{2}\times 0.224=1.456mol$ of oxygen gas

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

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

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water

So, 0.224 moles of butane will produce = $\frac{10}{2}\times 0.224=1.12moles$ of water

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

Molar mass of water = 18 g/mol

Moles of water = 1.12 moles

Putting values in equation 1, we get:

$1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol\times 18g/mol)=20.16g$

Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams

8 0

3 years ago