A neutral atom of potassium has 19 electrons.
Answer:
x = 4.17
y = 1.86
Explanation:
0.62 = log(x)
x = 10^0.62 = 4.17 ( to the nearest hundredth)
0.62 = ln(y)
y = e^0.62 = 1.86 (to the nearest hundredth)
Answer:
Percent yield = 84.5 %
Explanation:
Given data:
Mass of methanol = 229 g
Actual yield of water = 219 g
Percent yield of water = ?
Solution:
Chemical equation:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
Number of moles of methanol:
Number of moles = mass/ molar mass
Number of moles = 229 g/ 32 g/mol
Number of moles = 7.2 mol
Now we will compare the moles of water with methanol.
CH₃OH : H₂O
2 : 4
7.2 : 4/2×7.2 = 14.4 mol
Mass of water:
Mass = number of moles × molae mass
Mass = 14.4 mol × 18 g/mol
Mass = 259.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 219 g / 259.2 g × 100
Percent yield = 84.5 %
The balanced equation reaction :
2Cu+(aq) + PbO2(s) + SO42–(aq) + 4H+ -----------> PbSO4(s) + 2 Cu2+(aq) + 2H2O
A balanced equation is an equation for a chemical reaction in which the range of atoms for every detail in the reaction and the full charge is identical for both the reactants and the products. In other phrases, the mass and the fee are balanced in each aspect of the reaction.
A balanced equation happens while the variety of the atoms involved inside the reactants side is identical to the variety of atoms in the goods facet. in this chemical reaction, nitrogen (N2) reacts with hydrogen (H) to provide ammonia (NH3). The reactants are nitrogen and hydrogen, and the product is ammonia.
A chemical equation is the symbolic representation of a chemical reaction inside the form of symbols and formulae, wherein the reactant entities are given on the left-hand side and the product entities.
Eo = Eored - Eo oxd
= 1.69 - 0.153
= 1.54 V
Eocell is positive so the reaction is spontaneous.
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Answer:
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
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