Answer:
2.4 dm-3
Explanation:
Equation of the reaction;
2HCl(aq) + Ba(OH)2(aq) -------> BaCl2(aq) + 2 H2O(l)
Volume of acid VA = ??
Concentration of acid CA = 0.25 moldm-3
Volume of base VB = 2.0 dm^3
Concentration of base CB = 0.15 moldm-3
Number of moles of acid NA = 2
Number of moles of base NB = 1
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VA = CBVBNA/CANB
VA = 0.15 * 2 * 2/0.25 *1
VA = 2.4 dm-3
The radio wave travels at the speed of light so divide 8.00 x 10^7 by the speed of light.<span>
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For CO2 there is one atom of carbon and two atoms of oxygen. For H2O, there is one atom of oxyegen and two atoms of hydrogen. A moleclue can Be made of only one type of atom
Answer:
Density = 0.0035 g/cm³
Explanation:
The density equation looks like this:
Density (g/cm³) = mass (g) / volume (cm³)
You have been given both the mass and volume. However, the given volume is not in the correct units. To convert cm to cm³, you should simply raise the given volume to the power of 3.
24 cm = (24 cm)³ = 24³cm³ = 13,834 cm³
Now, you can plug you mass and volume into the equation and solve for density. The final answer should have 2 sig figs to reflect the given values.
Density = mass / volume
Density = 48 g / 13,824 cm³
Density = 0.00347 g/cm³
Density = 0.0035 g/cm³
The mass of the formed precipitate of AgCl in the reaction is 1.29 grams.
<h3>How do we find moles from molarity?</h3>
Moles (n) of any substance from molarity (M) will be calculated by using the below equation:
M = n/V, where
V = volume in L
Given chemical reaction is:
2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)
- Moles of CaCl₂ = 0.150M × 0.03L = 0.0045 moles
- Moles of AgNO₃ = 0.100M × 0.015L = 0.0015 moles
From the stoichiometry of the reaction, mole ratio of AgNO₃ to CaCl₂ is 2:1.
0.0015 moles of AgNO₃ = reacts with 1/2×0.0045 = 0.00075 moles of CaCl₂
Here CaCl₂ is the limiting reagent, and formation of precipitate depends on this only.
Again from the stoichiometry of the reaction:
0.0045 moles of CaCl₂ = produces 2(0.0045) = 0.009 moles of AgCl
Mass of 0.009 moles AgCl will be calculated as:
n = W/M, where
- W = required mass
- M = molar mass = 143.45 g/mol
W = (0.009)(143.45) = 1.29g
Hence required mass of precipitate is 1.29 grams.
To know more about moles & molarity, visit the below link:
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