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4 years ago

What's WO2 in empirical form?

Chemistry

1 answer:

ch4aika [34]4 years ago

8 0

Answer:

WO2:

Explanation:

If any element symbol in a formula has no subscript, implying a subscript of one, the formula is already empirical.

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How many moles of oxygen are in 1 mole of Fe(NO3)3?

goldfiish [28.3K]

There are 97.2 mol of O in 10.8 mol of Fe(NO3)3

7 0

3 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

4 years ago

Which statement best describes the result of meiosis?

Vesna [10]

D. reproductive cells uwu

3 0

3 years ago

2.00g of Copper wire was added to 50.0 mL of a 0.050M solution of Silver Nitrate. The reaction

Svet_ta [14]

Answer:

See explanation

Explanation:

Now, the equation of the reaction is;

Cu(s) + 2AgNO3(aq) --------> Cu(NO3)3 (aq) + 2Ag(s)

Number of moles of silver precipitate = 0.24/108 g/mol= 0.0022 moles

Number of moles of AgNO3 = 50/1000 * 0.050 = 0.0025 Moles

If 2 moles of AgNO3 yields 2 moles of Ag

Then 0.0025 moles of AgNO3 yields 0.0025 * 2/2 = 0.0025 moles of Ag

Number of moles of Cu = 2.00g/63.5 g/mol = 0.03 moles

If 1 mole of Cu yields 2 moles of Ag

0.03 moles of Cu yields 0.03 moles * 2/1 = 0.06 moles of Ag

Hence AgNO3 is the limiting reactant

Theoretical yield of Ag = 0.0025 moles of Ag * 108 = 0.27 g

Actual yield of Ag = 0.24 g

Percent yield = actual yield/theoretical yield * 100

Percent yield = 0.24/0.27 * 100

Percent yield = 88.8%

4 0

3 years ago

The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi

Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the length in millimetres

Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing

b) Convert millimetres to metres

Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing

6 0

3 years ago