Answer:
6.72M of HNO3
Explanation:
In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:
290.7mL + 350.0mL = 640.7mL
And you are diluting the solution:
640.7mL / 350.0mL = 1.8306 times
As the original concentration was 12.3M, the final concentration will be:
12.3M / 1.8306 =
<h3>6.72M of HNO3</h3>
Part 1 : Answer is only B substance is soluble in water.
In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.
Part 2 : Substance A is not soluble in water and substance B is soluble in water.
According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.
But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.
Answer:2kg
Explanation:
Mass =?
Acceleration = 3.0 m/s2
Force = 6.0N
Force = Mass x Acceleration
6 = Mass x 3
Mass =6/3 = 2Kg
Answer:
The quantity of heat lost by the surroundings is 258,5J
Explanation:
The dissolution of salt XY is endothermic because the water temperature decreased.
The total heat consumed by the dissolution process is:
4,184 J/g°C × (75,0 + 1,50 g) × 0,93°C = 297,7 J
This heat is consumed by the calorimeter and by the surroundings.
The heat consumed by the calorimeter is:
42,2 J/°C × (0,93°C) = 39,2 J
That means that the quantity of heat lost by the surroundings is:
297,7J - 39,2J = <em>258,5 J</em>
I hope it helps!
<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
