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Dmitrij [34]
3 years ago
5

They want a mixture of 150 grams of water at 60 degrees Celsius, using hot water at 90 degrees Celsius and cold water at 14 degr

ees Celsius, how many grams of water should be used per face?​
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

Explanation:

Here,

Final mass(m)=150g

Final Temperature(t)=60°C

Temperature of hot water(t1)=90C

Temperature of cold water(t2)=14C

Let the grams of water to be used per face be m1 grams and (150-m1) grams for hot and cold water respectively.

We know that,

Heat lost by hot body=Heat gained by cold body

or, m1 s1 Δt1=m2 s2 Δt2

or, m1 (t1-t)=(150-m1) (t-t2)  [∵s1=s2, SHC of water is taken to be same]

or, m1 (90-60)=(150-m1) (60-14)

or, 30m1=6900-46m1

or, 76m1=6900

∴m1=90.79g

(150-m1)=59.21g

Hence, 90.79g and 59.21 g hot and cold water must be used respectively.

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Molar mass of H₂ = 1.008 × 2 g/mol = 2.016 g/mol <span>
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol </span><span>
Molar mass of HI = (1.008 + 126.9) g/mol = 127.9 g/mol 

H₂(g) + I₂(g) → 2HI </span><span>
Mole ratio H₂ : I₂ : HI = 1 : 1 : 2 </span><span>

Then the initial number of moles of H₂ = (3.35 g) / (2.016 g/mol) = 1.662 mol </span><span>
Initial number of moles of I₂ = (50.75 g) / (253.8 g/mol) = 0.2000 mol < 1.662 mol </span><span>
Hence, I₂ is the limiting reactant (limiting reagent). </span><span>

Number of moles of I₂ reacted = 0.2000 mol </span><span>
Number of moles of HI reacted = (0.2000 mol) × 2 = 0.4000 mol 
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With a diameter that's 11 times larger than Earth's, _______ is the largest planet.
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System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas
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Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

F=G \frac{m_1 m_2}{r^2}

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

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F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}

system B has masses m and 2m separated by a distance 2r:

F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}

system C has masses 2m and 3m separated by a distance 2r:

F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}

system D has masses 4m and 5m separated by a distance 3r:

F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

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