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Does 65 grams of calcium chloride stand for the mass number of calcium chloride
Molar mass of sodium chloride is 58.44 , therefore , 65 g is not the mass number of sodium chloride
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How many moles of Cu are needed to react with 5.8 moles of AgNO3? Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
Moles of Cu=4.72moles
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How many grams of calcium hydroxide do you need to measure out to prepare 3.00 L of a 12.0% (m/v) calcium hydroxide solution?
3.00 L of a 12.0% (m/v) calcium hydroxide solution . 100 ml Solution Contain 12 g of Ca ( OH ) 2 . So , 3000 ml Solution...
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1)Prepare n-butanyl bromide from n-butane
2)Prepare Ethyl bromide from n-butane
3)Prepare 1-Butanol from n-butane
4)Prepare 2-Butene from n-butane
1) CH3-CH2-CH2-CH3 + Cl2 ---gt; CH3-CH2-CH2-CH2-Cl + HCl CH3-CH2-CH2-CH2-Cl + KOH (dry) --gt; CH3-CH2-CH2=CH2 CH3-CH2-CH2=CH2 + Br2/H2O2 --gt; CH3-CH2-CH2-CH2-Br CH3-CH2-CH2-CH2-Br + KOH (aq) --gt; CH3-CH2-CH2-CH2-OH + KBr CH3-CH2-CH2-CH2-OH + NaBr/H2SO4...
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how many grams of NaCl should be weighed to prepare 1.00L of 601 ppm solution of Na+? how many grams of NaCl should be weighed to prepare 1.00 L of XYZ ppm solution of Na+?
Quantity of NaCl =4.72g
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Calculate the moles of calcium chloride (CaCl2) needed to react in order to produce 85.00 grams of calcium carbonate (CaCO3).
The reaction between the mentioned substances can be shown as following:CaCl2 + Na2CO3 = 2NaCl + CaCO3First, the mass of CaCO3 should be converted into moles:n = m/Mrwhere n -...
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How many grams of copper are required to react with 45.0 grams of silver nitrate? Cu+ 2 AgNO3= Cu(NO3)2 + 2 Ag
Solution:The balanced chemical equation:Cu + 2AgNO3= Cu(NO3)2 + 2AgAccording to the equation: n(Cu) = n(AgNO3)/2Moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3The molar mass of AgNO3...
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Using the following chemical equation how many grams of calcium chloride can be produced by reacting 57.09 g of calcium hydroxide? 2HCl + Ca(OH)<sub>2</sub> -> 2H<sub>2</sub>O + CaCl<sub>2</sub>
2HCl + Ca(OH)2 -> 2H2O + CaCl2According to the equation, the mass of calcium chloride equals:m(CaCl2) = m(Ca(OH)2) × Mr(CaCl2) / Mr(Ca(OH)2) = 57.09 g × 110.98 g/mol / 74.09 g/mol...
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AgNO3+NH4Cl --> AgCl+NH4NO3. If you have 50.26g of AgNO3, how many grams of NH4NO3 can be produced?
AgNO3+NH4Cl = AgCl+NH4NO3 n(AgNO3) = n(NH4NO3) n(AgNO3) = 50,26 / 169,87 = 0,296 mol = n(NH4NO3) m(NH4NO3) = n(NH4NO3) * Mr(NH4NO3) m(NH4NO3) = 0,296 * 80 = 23,67 g
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According to the following reaction, how many grams of ammonium chloride are needed to form 30.8 grams of hydrochloric acid?
ammonium chloride (aq) ammonia (g) + hydrochloric acid (aq)
. m (HCl) =30.8 g .m (NH4Cl)=? The reaction is NH4Cl= NH3 + HCl Let's find the amount of substance (HCl)= m(HCl)/ Mr(HCl) Mr (HCl)= Ar(H)+ Ar(Cl)=1+35.5= 36.5 . n=30.8/36.5=0.84...
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