Answer:
Multiuse nail forms.
Explanation:
Hello,
Nails extensions have been widely used to provide nails an extra length which is supposed to enhance the nails appearance. In such a way, different types of materials have been enforced in the light of the creation of those extensions which are known as multiuse nail forms which are pre-shaped plastic or aluminum enhancements used as a guide to extend nail enhancements beyond the fingertip for additional length. Materials such as cyanoacrylate (derived from acrylates) and aluminium turn out suitable for the manufacture of those multiuse nail forms as they provide stability during nail enhancement practices.
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Answer:
The answer to your question is 29.7 g of CaCl₂
Explanation:
data
mass CaCl₂ = ?
mass CaCO₃ = 27 g
mass HCl = 12 g
Balanced Reaction
CaCO₃ + HCl ⇒ CaCl₂ + H₂O + CO₂
Molecular weight
CaCO₃ = 40 + 12 + 48 = 100 g
HCl = 36 + 1 = 37 g
- Calculate proportions
Theoretical proportion CaCO₃ / HCl = 100 / 37 = 2.7
Experimental proportion CaCO₃/ HCl = 27/12 = 2.25
From the proportions we determine that the limiting reactant is CaCO₃, because the proportion diminishes.
Molecular weight CaCl₂ = 40 + 35.5 = 110 g
100 g CaCO₃ -------------------- 110 g of CaCl₂
27 g CaCO₃ ------------------------- x
x = (27 x 110) / 100
x = 29.7 g of CaCl₂
b) ion
c) Free radical
d) None
Answer:
D
Explanation:
None of the following options are ionic compounds
Answer:
Cp= 1.005 kJ/kg °C = 1.005 kJ/(kg*K) = 1.005 J/g°C = 4.206 J/g°C = 0.776 BTU/lb°F
Explanation:
for the specific heat
1) Cp= 1.005 kJ/kg °C * (1 °C/ 1 K) (temperature differences) = 1.005 kJ/(kg*K)
2) Cp= 1.005 kJ/kg °C * (1000 J/ kJ)* (1 kg/1000 gr) = 1.005 J/g°C
3) Cp= 1.005 kJ/kg °C * (4.186 kcal/kJ) = 4.206 J/g°C
4) Cp= 1.005 kJ/kg °C * (1 BTU/1.055 kJ)* (0.4535 kg/lb)*(1.8°C/°F)= 0.776 BTU/lb°F
The correct option is D.
The major component of rust is Fe2O3. To know the number of atom in one molecule of rust, one have to look at the chemical formula of rust and determine the different atoms that are present and their numbers. Looking at the chemical formula above, one will notice that, two atoms of iron is present and three atoms of oxygen. The number of atoms present are written in front of the chemical symbol of each element. two atoms of iron added to three atoms of oxygen will give 5 atoms. Thus, five atoms are present in one molecule of rust.