374u
187u
C₁₄H₂₂N₄O₈
Explanation:
To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.
The empirical weight is determined by using the simplest ratio of the elements involved in the compound;
Molecular weight of C₁₄H₂₂N₄O₈;
atomic mass of C = 12g/mol
H = 1g/mol
N = 14g/mol
O = 16g/mol
Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)
= 168 + 22 + 56 + 128
= 374u
Empirical weight:
Empirical formula:
C₁₄ H₂₂ N₄ O₈
14 : 22 : 4 : 8
divide by 2:
7 : 11 : 2 : 4
empirical formula C₇H₁₁N₂O₄
empirical weight = 
= 
= 187u
The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈
learn more:
Molecular mass brainly.com/question/5546238
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C. dim light is <span>the answer
</span>
Answer:
2 C2H2 + 5 O ---> 4 CO2 + 2 H2O
Explanation:
combustion reactions always end with CO2 + H2O
and you can use this website to balance out equations when you're stuck
https://en.intl.chemicalaid.com/tools/equationbalancer.php?equation=C2H2+%2B+O2+%3D+CO2+%2B+H2O
Answer is: the specific heat capacity of the metal is <span>A) 0.129 J/gK.
</span>m(metal) = 15,1 g.
Q = 48,75 J.
ΔT = 25 K.
Q = C · ΔT · m(metal).
C = Q ÷ ΔT · m(metal).
C = 48,75 J ÷ 25 K · 15,1 g.
C = 0,129 J/g·K.
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:
brainly.com/question/24322641
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