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nalin [4]
2 years ago
12

Pls help i will mark brainlist

Chemistry
1 answer:
Veseljchak [2.6K]2 years ago
4 0

58.5 moles

can i plz get brainliest

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If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant
Vanyuwa [196]
<h3>Answer:</h3>

              Excess Reagent =  NBr₃

<h3>Solution:</h3>

The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,

                       2 NBr₃ + 3 NaOH   →   N₂ + 3 NaBr + 3 HBrO

Calculating the Limiting Reagent,

According to Balance equation,

               2 moles NBr₃ reacts with  =  3 moles of NaOH

So,

           40 moles of NBr₃ will react with  =  X moles of NaOH

Solving for X,

                       X  =  (40 mol × 3 mol) ÷ 2 mol

                       X  =  60 mol of NaOH

It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.

6 0
3 years ago
(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n
quester [9]

Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene =\frac{0.03 g}{78 g/mol}=0.0003846 mol

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

\frac{15}{5}\times 0.0003846 mol=0.0011538 mol of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

4 0
3 years ago
When designing an experiment, the first step is to. ______. Group of answer choices a. hypothesis b. list a procedure c. state t
True [87]

Answer:

A

Explanation:

It's the scientific method

8 0
3 years ago
What would the volume of gas be at 150 c if it had a volume of 693 ml at 45 c​
marissa [1.9K]

Answer:

\boxed{\text{922 mL}}

Explanation:

The pressure is constant, so we can use Charles' Law to calculate the volume.

\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}

Data:

V₁ = 693 mL; T₁ =  45 °C

V₂ = ?;           T₂ = 150 °C

Calculations:

(a) Convert temperature to kelvins

T₁ = (  45 + 273.15) = 318.15 K

T₂ = (150 + 273.15) = 423.15 K

(b) Calculate the volume

\dfrac{ 693}{318.15} = \dfrac{ V_{2}}{423.15}\\\\2.178 = \dfrac{ V_{2}}{423.15}\\\\V_{2} = 2.178 \times 423.15 = \boxed{\textbf{922 mL}}

5 0
3 years ago
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