You would set this up as Part÷Whole=Percent÷100
Part=117
Whole= X
Percent=78
117÷X=78÷100
Set these up as fractions and cross multiply. You will get 117×100=78X
117×100=11,700
Divide both sides by 78 and you get x=150
There were 150 problems

6 0

3 years ago

85.87 J or heat energy is added to a 34.8 g mass of substance the temperature rises from 21.76°C

Andru [333]

Answer:

The required specific heat is 196.94 joule per kg per °C

Step-by-step explanation:

Given as :

The heat generated = Q = 85.87 J

Mass of substance (m)= 34.8 gram = 0.0348 kg

Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C

Let the specific heat = S

Now we know that

Heat = Mass × specific heat × change in temperature

Or, Q = msΔt

Or, 85.87 = (0.0348 kg ) × S × 12.53°C

Or , 85.87 = 0.4360 × S

Or, S = $\frac{85.87}{0.4360}$

∴ S = 196.94 joule per kg per °C

Hence the required specific heat is 196.94 joule per kg per °C Answer

7 0

2 years ago

Find three rational number between -5/6and3/8​

Find three rational number between -5/6and3/8