Answer:
0.934J/g°C
Explanation:
Using Q = mc∆T
However, in this question;
(Q)water = -(Q)metal
(mc∆T)water = -(mc∆T)metal
According to the information provided in the question;
For water;
m = mass = 72.9g
c = specific capacity of water = 4.184 J/g°C
∆T = 22.9 - 15.9 = 7°C
For metal;
m = mass = 45.7g
c = specific capacity of water = ?
∆T = 22.9 - 72.9 = -50°C
(mc∆T)water = -(mc∆T)metal
(72.9 × 4.184 × 7) = -(45.7 × c × -50)
2135.0952 = -(-2285c)
2135.0952 = 2285c
c = 2135.0952/2285
c of metal = 0.934J/g°C
Answer:
Fluorine
Explanation:
Data obtained from the question include:
The isotope of fluorine has:
Proton = 9
Neutron = 10
The atomic number is the proton number of the atom.
Therefore, the atomic number = 9
Mass number = proton + Neutron
Mass number = 9 + 10
Mass number = 19
Now comparing the above with the fluorine atom in the periodic table. We'll discover that the isotope is actually fluorine atom.
Therefore, the name of the isotope is fluorine
A control group is the comparison group that helps to "make sure your experiment works." A control group is separated from the rest of the experiment and nothing happens to it kinda like a controlled variable. Controlled variables are the variables in a experiment that remains the same for example a temperature, time, type of products, etc..
Hope this helps!
Answer:
N2(g) + 3H2(g) → 2 NH3(g)
Explanation:
N2(g) + H2(g) → NH3(g)
We start equaling the number of N atoms in both sides multiplying by 2 the NH3.
N2(g) + H2(g) → 2 NH3(g)
So we equals the H atoms (there are six in products sites)
N2(g) + 3 H2(g) → 2 NH3(g)
