What are five objects that are luminous

Bronze is an example of which of the following: element atom solution solvent

Brut [27]

It is a solution, because it's the alloy of tin and copper

8 0

3 years ago

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

GaryK [48]

First we can calculate the area of the rectangular lawn using the formula:
Area = Width x Length = 21 ft x 20 ft = 420 square feet

And the total number of snow flakes per minute on the entire lawn is:

(1350 snowflakes per minute per square foot) x (420 square feet) = 567,000 snowflakes per minute

In one hour (or 60 minutes) we get a total of:

(567,000 snowflakes per minute) x (60 minutes / 1 hour) = 34,020,000 snowflakes

The total mass of which would be:

34,020,000 snowflakes x 1.60 mg = 54,432,000 mg = 54.432 kg (as 1 kg = 1,000,000 mg).

So 54.432 kg of snow accumulates every hour on the lawn.

7 0

3 years ago

Specific heat of Fe=0.473 J/g degree C; specific hear of Pb= 0.0305 J/g degree Celcius. To raise the temperature of 1.0 g of the

enot [183]

B. Fe requires more energy than does Pb.

8 0

3 years ago

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 (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), 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: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

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