Impossible to form from given reactants
Explanation:
Only sodium has carbonate ion here .so only possible way to displace carbonate is to use potassium
When sodium carbonate and Potassium iodide react with each other double displacement reaction occurs
As potassium is powerful than sodium it displaces sodium to form potassium carbonate .
Then you have to use water and then you can use calcium suppliment .
MFe: 56 g/mol
..........
56g ----- 1 mol
5g ------- X
X = 5/56 = 0,089 moles of Fe
56g ----- 6,02×10²³ atoms
5g ------- X
X = (5×6,02×10²³)/56
X = 0,5375×10²³ = 5,375×10²² atoms of Fe
:•)
Answer:
Mass of KNO3 in the original mix is 146.954 g
Explanation:
mass of 
in original 254.5 mixture.
moles of 
moles of
= 0.2926 mol of BaSO4
Therefore,
0.2926 mol of BaCl2,
mass of 
= 60.92 g
the AgCl moles 
= 1.3891 mol of AgCl
note that, the Cl- derive from both, 
so
mole of Cl- f NaCl 
mol of Cl-
mol of NaCl = 0.8039 moles
then
KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3
Mass of KNO3 in the original mix is 146.954 g
Answer :]
A.)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.
<em>Correct me if i'm wrong :]</em>
The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)
<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
- Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–38) = 38 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.4 × 2050 × 38
Q₁ = 31160 J
<h3>How to determine the heat required to melt the ice at 0 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.4 × 334000
Q₂ = 133600 J
<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 100 °C
- Change in temperature (ΔT) = 100 – 0 = 100 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.4 × 4180 × 100
Q₃ = 167200 J
<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
- Heat (Q₄) =?
Q = mHv
Q₄ = 0.4 × 2260000
Q₄ = 904000 J
<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 100 °C
- Final temperature (T₂) = 160 °C
- Change in temperature (ΔT) = 160 – 100 = 60 °C
- Specific heat capacity (C) = 1996 J/(kg·°C)
- Heat (Q₅) =?
Q = MCΔT
Q₅ = 0.4 × 1996 × 60
Q₅ = 47904 J
<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
- Heat for –38 °C to 0°C (Q₁) = 31160 J
- Heat for melting (Q₂) = 133600 J
- Heat for 0 °C to 100 °C (Q₃) = 167200 J
- Heat for vaporization (Q₄) = 904000 J
- Heat for 100 °C to 160 °C (Q₅) = 47904 J
- Heat for –38 °C to 160 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qₜ = 31160 + 133600 + 167200 + 904000 + 47904
Qₜ = 1.28×10⁶ J
Learn more about heat transfer:
brainly.com/question/10286596
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