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Vlad1618 [11]
2 years ago
15

1.000 mile is equal to 1609 meters. A truck is traveling at 42.00 km/h. What is the speed of the truck in miles per hour?

Chemistry
2 answers:
schepotkina [342]2 years ago
5 0

Answer : The speed of truck in miles per hour is 26.103 mile/hr

Explanation :

The conversion used from kilometer to meter is:

1 km = 1000 m

The conversion used from meter to miles is:

1 mile = 1609 m

or,

1m=\frac{1}{1609}mile

and, 1km=\frac{1000m}{1609m}\times 1mile

So, 1km/hr=\frac{1000}{1609}\times 1mile/hr

As we are given that the speed of truck is 42.00 km/hr. Now we have to calculate the speed of truck in miles per hour.

As, 1km/hr=\frac{1000}{1609}\times 1mile/hr

So, 42.00km/hr=\frac{42.00km/hr}{1km/hr}\times \frac{1000}{1609}\times 1mile/hr

                      = 26.103 mile/hr

Therefore, the speed of truck in miles per hour is 26.103 mile/hr

bija089 [108]2 years ago
3 0

Around 24.85 miles per hour

, divide the length value by 1.60

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-BARSIC- [3]
Molar mass of sulfur dioxide (SO2) is 64.066g/mol
molar mass of sulfur is 32.065 g/mol
thus, (32.065/64.066 )x100=50% 

7 0
2 years ago
A heat energy of 645 J is applied to a sample of glass with a mass of 28.4 g. Its temperature increases from –11.6 ∞C to 15.5 ∞C
Anika [276]
The heat that is required to raise the temperature of an object is calculated through the equation,
                        heat = mass x specific heat x (T2 - T1)
Specific heat is therefore calculated through the equation below,
                                specific heat = heat / (mass x (T2 - T1))
Substituting,
                                specific heat = 645 J / ((28.4 g)(15.5 - - 11.6))
The value of specific heat from above equation is 0.838 J/g°C. 
5 0
2 years ago
0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

8 0
2 years ago
calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.
Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

P_w=P_p+P_s

P_w=P_p+(-iCRT)

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= P_w

P_p=0 bar (At standard temperature)

P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar

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The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

7 0
2 years ago
A patient needs to be given exactly 500 ml of a 5.0% (w/v) intravenous glucose solution. the stock solution is 35% (w/v). how ma
N76 [4]
In this question, the <span>patient needs to be given exactly 500 ml of a 5.0%. The content of the glucose should be:
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500ml * 1mg/ml *5%= 25mg.

The </span><span>stock solution is 35%, then the amount needed in ml would be: 
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25mg= volume * 1mg/ml *35%
volume= 25/35%= 500/7= 71.43ml</span>
6 0
2 years ago
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