Answer:
0.0258 mol <em>Answer</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
Protons: 79
Electrons: 78
Explanation:
1. The number of protons is the atomic number (The atomic number for Au on the periodic table is 79)
2. Since the charge is +1 (positive) it means that there's one more proton than electrons. So, 79-1 = 78 electrons
Answer:
A) positive; added
Explanation:
Based on the reaction:
2NaHCO3(s) + 129kJ → Na2CO3(s) + H2(g) + CO2(g)
<em>2 moles of NaHCO3 requires 129kJ to produce 1 mole of Na2CO3, 1 mole of H2 and 1 mole of CO2.</em>
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That means, the energy must be added being, thus, an exothermic reaction. The exothermic reactions have ΔH >0.
Thus, right answer is:
A) positive; added
Answer: 483 mL of the cleaning solution are used to clean hospital equipment
Explanation:
The question requires us to calculate the volume, in mL, of solution is used to clean hospital equipment, given that 415g of this solution are used and the specific gravity of the solution is 0.860.
Measurements > Density
Specific gravity is defined as the ratio between the density of a given substance to the density of a reference material, such as water:

The density of a substance is defined as the ratio between the mass and the volume of this substance:

Considering the reference substance as water and its density as 1.00 g/mL, we can determine the density of the substance which specific gravity is 0.860:

Thus, taking water as the reference substance, we can say that the density of the cleaning solution is 0.860 g/mL.
Now that we know the density of the cleaning solution (0.860 g/mL) and the mass of solution that is used to clean hospital equipment (415g), we can calculate the volume of solution that is used to clean the equipment:

Therefore, 483 mL of the cleaning solution are used to clean hospital equipment.
Answer:
ΔH = 57.04 Kj/mole H₂O
Explanation:
60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)
=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)
=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)
=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat
ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O
=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)
ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O