In chemistry, if you want to express the amount of a substance out of the total amount, you express it in concentration. There are numerous units of measurement: molarity, molality, normality, mass percentages, volume percentage, or a mix of both. For this problem, the unit used for concentration is in mass percentages. The formula would be
Percentage Concentration = [(Actual Amount of Substance)/(Total amount of all substances)] * 100
Since we are given with the total mass of all the substances in the ocean and the percentage concentration, the only missing information is the actual amount of Na+ in the ocean. Substituting the values:
1.076 = (Amount of Na+ /1.8×10²¹ kg)*100
Amount of Na+ = 1.9368×10¹⁹ kg
Answer: put 2 in front of NaOH and 2 also in front of H2O
Explanation: 2NaOH + H2CO3 —> Na2CO3 + 2H2O
Answer:
If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial
In this case:
- Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)
- m=700 g
- ΔT= Tfinal - Tinitial= Tfinal - 90 °C
Replacing:
Solving:
16.125 °C= Tfinal - 90 °C
Tfinal= 16.125 °C + 90 °C
Tfinal= 106.125 °C
<u><em>If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C</em></u>
Answer:
16.9g of H₂O can be formed
Explanation:
Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
8.76g * (1mol / 2.016g) = 4.345 moles
<em>Moles O₂:</em>
PV = nRT
PV/RT = n
P = 1atm at STP
V = 10.5L
R = 0.082atmL/molK
T = 273.15K at STP
n = 1atm*10.5L / 0.082atmL/molK*273.15K
n = 0.469 moles of oxygen
For a complete reaction of 4.345 moles moles of hydrogen are required:
4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant
Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:
0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.
The mass is -Molar mas H₂O = 18.01g/mol-:
0.938 moles * (18.01g/mol) =
<h3>16.9g of H₂O can be formed</h3>
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