Answer:
This metal could be the aluminium with a specific heat of 
Explanation:
A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows
We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that
Finally, the specific heat of this metal is
The aluminium could be the metal, its specific heat is similar to that found in this problem.
Finally, we can conclude that this metal could be the aluminium with a specific heat of
Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol
1 mole ---------- 84 g
? mole ---------- 110 g
moles NaHCO₃ = 110 . 1 / 84
moles NaHCO₃ = 110 / 84
= 1.309 moles
hope this helps!
Answer:
in this the correct answer is option 2.
Answer:
1) d = 2.4 g/cm³
2) m = 25 g
3) v = 126.7 cm³
Explanation:
Given data:
Mass of material = 24 g
Volume of material = 10 cm³
Density of material = ?
Solution:
Formula:
d = m/v
by putting value,
d = 24 g / 10 cm³
d = 2.4 g/cm³
2) Given data:
Density of material = 5 g/cm³
Volume of material = 5 cm³
Mass of material = ?
Solution:
Formula:
d = m/v
5 g/cm³ = m / 5 cm³
m = 5 g/cm³×5 cm³
m = 25 g
3)Given data:
Density of material = 3 g/cm³
Mass of material = 380 g
Volume of material = ?
Solution:
Formula:
d = m/v
3 g/cm³ = 380 g / v
v = 380 g /3 g/cm³
v = 126.7 cm³
Answer:
The answer is "
"
Explanation:
Please find the complete question in the attached file.
Equation:
at 
at equilibrium 
![p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\](https://tex.z-dn.net/?f=p%3D%200.47%20%5C%20%5C%20atm%5C%5C%5C%5CSO_2%3D3.3-0.47%20%3D%202.83%20%5C%20%5C%20atm%5C%5C%5C%5CO_2%3D%200.74%20-%5Cfrac%7B0.47%7D%7B2%7D%3D0.74-0.235%3D0.555%20%5C%20atm%5C%5C%5C%5CK_P%3D%5Cfrac%7B%5BPSO_3%5D%5E2%7D%7B%5BPSO_2%5D%5E2%5BPO_2%5D%7D%5C%5C%5C%5C)
