Answer:
2L of water.
Explanation:
To know the volume of water to be added to the initial solution, first let us calculate the volume of the final solution. This is illustrated below:
Data obtained from the question:
Initial volume (V1) = 2L
Initial concentration (C1) = 6mol/L
Final concentration (C2) = 3mol/L
Final volume (V2) =?
Using the dilution formula, we can obtain the final volume of the stock as follow:
C1V1 =C2V2
6 x 2 = 3 x V2
Divide both side by 3
V2 = (6 x 2)/3
V2 = 4L.
The final volume of the solution is 4L.
To obtain the volume of water added, we shall determine the change in the volume of the solution. This is illustrated below:
Initial volume (V1) = 2L
Final volume (V2) = 4L
Change in volume = V2 – V1 = 4 – 2 = 2L.
Therefore, 2L of water must be added to the initial solution.
Answer: 10 moles
To find how many moles, you need to divide the number of atoms with 6.02 * 10^23. Since every magnesium chloride has two chlorine atoms, you need to multiply it by 2. The calculation would be: 3.01×10^24 * 2 / 6.02 * 10^23= 10 moles
Explanation:
1kg x 0,37=370gHCL
370g/36,5g/mol=10,137mol/kg
1l=1,185kg
10,137 x 1,185kg/l =12,0mol/l
(x/ml(12mol HCL)*12mol/L)/100ml=0,025mol/L
x=
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
The organism would no longer grow.
