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gulaghasi [49]
3 years ago
8

A horse was 6,200 meters north of a barn. For 22.5 minutes, the horse ran toward the barn at 152 meters per minute. In that time

, how far did the horse run?
Chemistry
1 answer:
lutik1710 [3]3 years ago
6 0

3420‬m

Explanation:

Given parameters:

Length of path = 6200m

time taken for the horse to run = 22.5min

speed = 152meters/minute

Unknown:

Distance run by the horse = ?

Solution:

Speed is the rate of change of distance with time.

   Speed       =    \frac{distance}{time}

    Distance = speed x time

 

     Distance = 152 x 22.5 = 3420‬m

learn more:

Speed          brainly.com/question/5424148

#learnwithBrainly

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Answer: covalent bond is the force of attraction that holds together two atoms that share a pair of valence electrons. Covalent bonds form only between atoms of nonmetals. The two atoms that are held together in a covalent bond may be atoms of the same element or different elements.

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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh
Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

Molarity = M = \frac{1,22 mol solute}{lts solution}

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent  

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03

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If a sample of air in a closed container was heated, the total pressure of the air would increase.

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Answer:

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