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otez555 [7]
2 years ago
6

How do human beings prepare their food​

Chemistry
2 answers:
Rudiy272 years ago
3 0
Well they buy food then they cook it then they eat it
Mademuasel [1]2 years ago
3 0

Answer:

sugar and spice and everything nice

Explanation:

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HLP PLZ. explain why you need to use more than one property to identify a mineral.
marin [14]

Answer:

Because  two minerals can share property's so you need more then one to tell them apart. Hope this helped! :)

Explanation:

8 0
3 years ago
How many moles of C2H2 are needed to react completely with 84.0 mol O2?
Airida [17]
33.6 moles are needed to completely react with 84.0 moles of O2
3 0
2 years ago
Read 2 more answers
When bismuth-212 undergoes alpha decay, it becomes which of the following?
Iteru [2.4K]

Answer: -

When bismuth-212 undergoes alpha decay, it becomes ²⁰⁸Tl

Explanation: -

Mass number of ²¹²Bi = 212

Atomic number of ²¹²Bi = 83

When alpha decay occurs the mass number decreases by 4 and the atomic number decreases by 2.

Mass number of daughter = 212 - 4 = 208

Atomic number of daughter = 83 - 2 = 81

It is the atomic number of Thallium Tl.

Thus the daughter nucleide is ²⁰⁸Tl.

8 0
2 years ago
Read 2 more answers
A short-lived structure formed during a collision is a(n)
Jet001 [13]
Beef and cheddar I believe!!!
3 0
3 years ago
Read 2 more answers
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
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