Answer:
Explanation:
has a pH of 6.6, then what is the H3O+ in solution X? View Answer · What is the pOH of a solution in which (H+)
Answer:
The answer to your question is below
Explanation:
Data
mass of CaCO₃ = 155 g
mass of HCl = 250 g
mass of CaCl₂ = 142 g
reactants = CaCO₃ + HCl
products = CaCl₂ + CO₂ + H₂O
1.- Balanced chemical reaction
CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O
2.- Limiting reactant
molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
molar mass of HCl = 2[1 + 35.5 ] = 73 g
theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37
experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃
3.-
Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 111 g
100 g of CaCO₃ ------------------ 111 g of CaCl₂
155 g of CaCO₃ ----------------- x
x = (155 x 111) / 100
x = 17205 / 100
x = 172.05 g of CaCl₂
4.- percent yield
Percent yield = 142 / 172.05 x 100 = 82.5 %
5.- Excess reactant
100 g of CaCO₃ -------------------- 73 g of HCl
155 g of caCO₃ ------------------- x
x = (155 x 73)/100
x = 133.15 g
Mass of HCl = 250 - 133.15
= 136.9 g
Answer:
EIt involves copying a gene's DNA sequence to make an RNA molecule. Transcription is performed by enzymes called RNA polymerases, which link nucleotides to form an RNA strand (using a DNA strand as a template). ... Transcription is controlled separately for each gene in your genome.xplanation:
Answer:
<em>Ionic</em><em> </em>
If something is a hard solid, that can dissolve in water, and has a high melting point are ionic substances, because they are hard solids due to strong electrostatic force between particles and and they have high melting points because of their strong attraction between the particles which needed more energy in order to break down and melt .....
their one property is to dissolve in water because they are generally non polar which dissolves easily in polar liquids like water.
<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em>
