Answer:
Explanation:
Assume you are using 1 L of water.
Then you are washing 4 L of salty oil.
1. Calculate the mass of the salty oil
Assume the oil has a density of 0.86 g/mL.
2. Calculate the mass of salt in the salty oil
3. Calculate the mass of salt in the spent water
4. Mass of salt remaining in washed oil
Mass = 172 g - 150 g = 22 g
5. Concentration of salt in washed oil
Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
Answer:
D.
The concentration of reactants and the concentration of products are constant.
Explanation:
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Answer:
2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3
Explanation:
First of all, let us balance the equation to give;
2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 3NaOH
Now, we can observe the presence of positive Sodium ions (Na+) and negative hydroxyl ions (OH-) on both left and right sides of the equation.
Now, the two ions will cancel out. These ions are not really involved in the overall reaction and thus do not require being written in the overall equation. Hence, the overall net ionic reaction can now be written as:
2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
