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crimeas [40]
3 years ago
6

What is the concentration in molarity of a solution made using 50.0 grams of C6H12O6 in 300.0 mL of water?

Chemistry
2 answers:
oee [108]3 years ago
7 0

Answer: concentration in Molarity it's actually Molarity

Molarity=concentration/molar mass

Concentration=50g/0.3dm³=166.67g/dm³

Molar mass=180g/mol

Molarity=0.926mol/dm³

Sladkaya [172]3 years ago
7 0

Answer:

Answer is =0. 925 mol L−1

Explanation:

C₆H₁₂O₆ is formula of Glucose  

For Molarity, we must know the following things  

• the number of moles of solute present in solution

• the total volume of the solution

we know the mass of one mole of Glucose = 180.156 g/mol

Number of moles = Given Mass of substance / Mass of one mole

No of moles = 50g / 180.156 g/mol

                   = 0.277 moles  

Now we know that  molarity is expressed per liter of solution. Since you dissolve 0.277 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain

For 300 ml of solution, no of moles are = 0.277 moles

For 1 ml of solution, no of moles are = 0.277/300 moles

For 1(1000) ml of solution, no of moles are=  0. 277/300 x 1000

                                                                 = 0.925 moles/ L

Answer is =0. 925 mol L−1

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A steel beam that is 7.00 m long weighs 326 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
GrogVix [38]

Answer:

The answer to the question is

Suki is 1.12 m close to the end of the steel beam before the beam begins to tip

Explanation:

To solve the question we list out the known variables first

Length of steel beam, L = 7.00 m

Weight  of steal beam W = 326 N

Distance between supports = 3.00 m

Weight of Suki = 555 N

By moments theory, the sum of moments about a point = 0

and assuming the steal beam is uniform, the weight of the beam will act at the center of the beam

As described the supports are 2.00 m from each end of the beam thus

taking moments about one of the support with Suki between  the end of the beam and the support we have

Distance from center of beam to support = 1.5 m

Distance of Suki to the nearest support = x m

therefore 555 × x + 326 × 1.5 = 0

or x = -0.881 m on the other side of the support

Therefore Suki is (2 - 0.881) m or 1.12 m close to the end before the beam begins to tip

7 0
3 years ago
Acids or bases can be tested by chemical_________
DedPeter [7]

Answer:

acids or bases can be tested

by chemical indicators

7 0
2 years ago
Read 2 more answers
1. Which process uses oxygen?photosynthesis or cellular respiration
sleet_krkn [62]
1 Cellular Respiration
2 Photosynthesis 
3 Photosynthesis
4 Cellular respiration

5 0
3 years ago
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What happens on a molecular level when a diatomic molecule is a gas but is then cooled to a solid?
irina1246 [14]

Answer:

As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass.

Explanation:

Hopefully that helps!

4 0
3 years ago
The heat of fusion AH, of ethyl acetate (C4H802) is 10.5 kinol. Calculate the change in entropy as when 398. g of ethy, acetate
Hitman42 [59]

<u>Answer:</u> The entropy change of the ethyl acetate is 133. J/K

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethyl acetate = 398 g

Molar mass of ethyl acetate = 88.11 g/mol

Putting values in above equation, we get:

\text{Moles of ethyl acetate}=\frac{398g}{88.11g/mol}=4.52mol

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{fusion}}{T}

where,  

\Delta S = Entropy change  = ?

n = moles of ethyl acetate = 4.52 moles

\Delta H_{fusion} = enthalpy of fusion = 10.5 kJ/mol = 10500 J/mol   (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = 84.0^oC=[84+273]K=357K

Putting values in above equation, we get:

\Delta S=\frac{4.52mol\times 10500J/mol}{357K}\\\\\Delta S=132.9J/K

Hence, the entropy change of the ethyl acetate is 133. J/K

7 0
3 years ago
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