Answer:
Specific heat of alloy = 0.2 j/ g.°C
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Given data:
Mass of bold = 25 g
Heat absorbed = 250 J
Initial Temperature = 25°C
Final temperature = 78°C
Specific heat of alloy = ?
Solution:
Change in temperature:
ΔT = 78°C - 25°C
ΔT = 53°C
Now we will put the values in formula.
Q = m.c. ΔT
250 j = 25 g × c ×53°C
250 j = 1325 g.°C × c
250 j / 1325 g.°C = c
c = 0.2 j/ g.°C
I think it's Almond Soy Milk because they're recommending your body's pH to be at 7.5 and the Almond Soy Milk is the answer with the closest pH to 7.5
Answer:
1: At temperatures below 542.55 K
2: At temperatures above 660 K
Explanation:
Hello there!
In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:
Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:
1:
It means that at temperatures lower than 542.55 K the reaction will be spontaneous.
2:
It means that at temperatures higher than 660 K the reaction will be spontaneous.
Best regards!
Answer:
Six electrons.
Explanation:
A carbon atom has six protons, so it must have six electrons.
Answer:
1. 31.25 mL
2. 1.98 g/L
3. 0.45 g/mL
Explanation:
For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.
1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.
(50.0 g)/(1.60 g/mL) = 31.25 mL
2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.
(0.891 g)/(0.450 L) = 1.98 g/L
3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.
(10.0 g)/(22.0 mL) = 0.45 g/mL
