Answer:Specific heat capacity at Constant volume of Nitrogen =742.32JKg.K
Explanation:
The Specific heat capacity at Constant volume of an ideal gas is given as
c =Cv / M
Cv= Constant volue of gas
M= Molar mass
But First, we determine the Constant volume, Cv which is given as
Cv = 5/2R since Nitrogen is a diatomic gas, N2 where R= 8.314Jmol.k
= 5/2 x 8.314Jmol.K
=20.785Jmol.K
Specific heat capacity at Constant volume, c
c= Cv/M
M=molar mass = 28.0 g/mol.
changing to kg/mol =28/1000= 0.028kg/mol
Therefore c =20.785Jmol.K/0.028kg/mol
=742.32JKg.K
I’ve only had one and it didn’t flow so ion think so
Answer:Hope this helps!
Explanation:
You can use a flame test to help identify the composition of a sample. The test is used to identify metal ions (and certain other ions) based on the characteristic emission spectrum of the elements. The test is performed by dipping a wire or wooden splint into a sample solution or coating it with the powdered metal salt. The color of a gas flame is observed as the sample is heated. If a wooden splint is used, it's necessary to wave the sample through the flame to avoid setting the wood on fire. The color of the flame is compared against the flame colors known to be associated with the metals.
Answer:
A 12.630 bar of chocolate contains 8315 milligrams of sugar
Explanation:
From the basics of conversion, we can find that:
1 gram is equal to 1000 mg
Using cross multiplication, we can find how many milligrams are present in 8.315 grams as follows:
1 gram .............> 1000 mg
8.315 grams ....> ?? mg
amount in milligrams = (8.315*1000) / 1 = 8315 milligrams
Hope this helps :)
The graphics in the attachment is part of the question, which was incomplete.
Answer: Fr = 102N and angle of approximately 11°.
Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:
Fx = 70+40-10 = 100
Fy = 40-20 = 20
Now, as the forces form a triangle, the totalforce is:
Fr = 
Fr = 
Fr = ≈ 102N
To determine the angle requested, we use:
arctg H = 
arctg H = 
H = tg 0.2 ≈ 11°.
