The product will not be affected by the addition of twice as much Na₂CO₃.
<h3>What is Limiting reagent in stoichiometry ?</h3>
- The maximum quantity of the end product determined by a balanced chemical equation is known as the Stoichiometry.
- The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced, and the one which remains unconsumed after the final reaction is in Excess.
- Calculate the moles of each reactant present and contrast it with the mole ratio of the reactants in the balanced equation to determine which reactant is the limiting one.
Here,taking the stoichiometry into consideration, we find that the reaction happens with 1:1 ratio; so, adding twice the amount of Na₂CO₃ will lead to its excess making the other the limiting reactant, hence, it would not affect the yield of the product.
To know more about the Limiting reactant, refer to:
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<span>With any element, there are a range of isotopes . . . a weighted average mass is used because there exist different isotopes of that element, each of varying mass. A weighted average accounts for the fact that the average mass (that may likely be measured from a random sample of the element) will most likely reflect the average mass of the more abundant isotopes. So the weighted average mass will be most comparable to the more abundant isotopes.</span>
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped :D
This is one of the ideal gas laws. Presumably the pressure remains the same so it is not part of the givens.
Formula
V / T = V1 / T1
Givens
- V = 56.05 mL
- T = 315.1 degrees Kelvin
- V1 = x
- T1 = 380.5 degrees Kelvin
Solution
56.05/315.1 = x/380.5 Simplify the left.
0.1779 = x / 380.5 Multiply both sides by 380.5
0.1779 * 380.5 = 380.5x/380.5
67.68 mL = x This is your answer
Answer
Q=4479.8 cal
Procedure
To solve the problem you will need to use the specific heat formula
Where;
Q=heat energy
m=mass
c=specific heat capacity
ΔT=change in temperature
Assuming that the heat released from the cracker of unknown material is equal to the heat absorbed by the water, then we can use the c and m for water in our calculations.
Substituting the values in our equation we have
Finally, transform the J to cal
