Electrons and newtrons is that what ur asking for
Answer:
84.4g of AgCl
Explanation:
Based on the reaction:
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
<em>2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca(NO₃)₂</em>
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100g of each reactant are:
AgNO₃: 100g × (1mol / 169.87g) = 0.589 moles
CaCl₂: 100g × (1mol / 110.98g) = 0.901 moles
For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901×2 = <em>1.802 moles of AgNO₃. </em>As there are just 0.589moles, <em>AgNO₃ is limitng reactant</em>
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0.589 moles of AgNO₃ produce:
0.589 moles × ( 2 moles AgCl / 2 moles AgNO₃) =
<em>0.589 moles of AgC</em>l. In mass:
0.589 moles of AgCl × (143.32g / mol) =<em> 84.4g of AgCl</em>
Answer:
D. It contains a phosphate with higher phosphoryl transfer potential than ATP
Explanation:
1,3-Bisphosphoglycerate contains a phosphate group that has high phosphoryl transfer potential than ATP (they can transfer the phosphoryl group to ATP). Other high phosphoryl transfer potential groups include :Creatine kinase and phosphoenolpyruvate.
Grams of Chromium(III) nitrate produced : 268.95 g
<h3>Further explanation</h3>
Given
0.85 moles of Lead(IV) nitrate
Required
grams of Chromium(III) nitrate
Solution
Reaction
Balanced equation :
<em>2Cr₂ + 3Pb(NO₃)₄ ⇒ 4Cr(NO₃)₃ + 3Pb </em>
From the equation, mol ratio of Pb(NO₃)₄ : Cr(NO₃)₃ = 3 : 4, so mol Cr(NO₃)₃
mol Cr(NO₃)₃ =

Mass of Chromium(III) nitrate (MW=238.0108 g/mol) :
mass = mol x MW
mass = 1.13 x 238.0108
mass = 268.95 g