Given is the specific heat of water equal to 4.18 Joule per gram per *C.
This means to raise the temperature of 1 g of water by 1 degree Celsius we need 4.18 joule of energy.
Now, look at the question. We are asked that how much amount of energy would be required to raise the temperature of 25 g of water by (54-50) = 4 degree celsius.
To do so we have formula
Q = m C (temperature difference)
Have a look at pic for answer
Answer:
The section of the bar is 2.92 inches.
Explanation:
Mass of the steel cut ,m = 1.00 kg = 1000 g
Volume of the steel bar = V = Area × height
Height of the of the section of bar = h
Area of Equilateral triangular = 
a = 2.50 inches
Cross sectional area of the steel mass = A
Density of the steel = d =
h = 2.92 inches
The section of the bar is 2.92 inches.
Answer:
Na2O+H2O=2NaOH
Step by step exp.
Given:
Equation Na2O+H2O=NaOH
To find: Balance the equation
Solution:
Taking LHS of the equation
LHS=Na2O+H2O
There is 2 sodium, 2 oxygen,& 2 hydrogen
To balance the equation we have equal number of atom so we multply 2 to the RHS=2NaOH
There fore the equation form is
Na2O+H2O=2NaOH
This
can be solved using Dalton's Law of Partial pressures. This law states that the
total pressure exerted by a gas mixture is equal to the sum of the partial
pressure of each gas in the mixture as if it exist alone in a container. In
order to solve, we need the partial pressures of the gases given. Calculations
are as follows:<span>
<span>P = 3.00 atm + 1.80 atm + 0.29 atm + 0.18 atm + 0.10 atm</span></span>
<span><span>P = 5.37 atm</span></span>
<span>
</span>
<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3
So, for 2.3 ✕ 10−6 m OH-
[Cu2+] =
