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Lemur [1.5K]
3 years ago
11

A group of tissues that work together to perform a specific function is called

Chemistry
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

B. Organ.

Explanation:

College Biology Major here.

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Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

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

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

  • <u>For B_5H_9 :</u>

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

  • <u>For oxygen gas:</u>

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
3 years ago
Thea is doing a chemistry experiment. The instructions say she needs to use pure water. Hamza offer her a bottle labelled “100%
Diano4ka-milaya [45]
How is this going to help us in life wat the fawk
5 0
3 years ago
HELP. I DONT KNOW WHAT TO DO. Barium sulfate, BaSO4, is made by the following reaction.
Ghella [55]

Answer:

\boxed{\text{66.95 g BaSO$_{4}$}}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:         261.34                         233.39

              Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃

m/g:         75.00

1. Moles of Ba(NO₃)₂

\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}

2. Moles of BaSO₄

The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂

\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}

3. Mass of BaSO₄

\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}

8 0
3 years ago
How many chromosomes are in a human gamete ?
RideAnS [48]

Answer:

23

Explanation:

if if you look at the screenshot it has the explanation

5 0
2 years ago
When magnesium reacts with chlorine, the chlorine atom gain electrons. what happens to chlorine in this reaction?
Mrrafil [7]

Answer:

D

Explanation:

Reduction refers to gain in electrons. Hence it is D

Oxidation occurs when the element loses electrons. Hence A is wrong

7 0
3 years ago
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