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2 years ago

What can do ,when the solution remain in the jet/ nozzle of the pipette?

Chemistry

1 answer:

lutik1710 [3]2 years ago

5 0

ans is opt (b)

Explanation:

the pippet reading is already adjusted accordingly so there is no need to blpw out the last drop ylthe reading will be accurate anyway

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Which can not be decomposed by a chemical change? 1) a compound 2) a heterogeneous mixture 3) a homogeneous mixture 4) an elemen

Mnenie [13.5K]

4.) an element cannot be decomposed by a chemical change, it is a pure substance and no other materials compose it.

3 0

3 years ago

Name the two molecules Nн нннн

arlik [135]

Answer:

1: C5H12

2:C5H11

Explanation:

nakqkchlqosnx

4 0

2 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

3 years ago

A gas is at 35.0°c and 3.50 l. what is the temperature at 7.00 l?

Degger [83]

Answer:

363C

Explanation:

V = KT and T = V/K

3.5=k(35+273) = 308k

k=3.5/308 =0.011

new T=7.0/0.011=636K = 636-273 =363C

4 0

2 years ago

What does the symbol (s) represent when it is next to an atom?

AlekseyPX

Answer:

The symbol for an atom indicates the element via its usual two letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript.

Explanation:

I hope that helps!! sorry if it dont!

8 0

2 years ago

What can do ,when the solution remain in the jet/ nozzle of the pipette?​

What can do ,when the solution remain in the jet/ nozzle of the pipette?