Answer:
1.034 L
Explanation:
P1 V1 = P2 V2
P1 V1 / P2 = V2
2.4 (2.5) / 5.8 = V2 = 1.034 L
First, find out how many grams are in one mole of CO2(the two oxygen atoms means you need to multiply oxygen’s amu by 2,then add whatever carbon’s amu is to that). Then divide 26 grams by that number and that will be your moles. There are only two significant figures, so round your answer correctly.
In order for you to get the answer, please have in mind the following situation: To increase the molar concentration of N2O4(g), 2NO2(g) should also increase for equilibrium to occur. Now, this equation is exothermic. By <span>Le Chatelier's principle, equilibrium constant and reaction constants also come into play in terms of increasing or decreasing the temperature. After that I know you can find the answer. </span>
91.4 grams
91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
C = mol/volume
2.45M=mol/0.5L
2.45M⋅0.5L = mol
mol = 1.225
Convert no. of moles to grams using the atomic mass of K + Cl
1.225mol * 
mol=1.225
=1.225 mol . 
=1.225 . 74.6
=91.4g
therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.
What is 1 molar solution?
In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.
58.44 g make up a 1M solution of NaCl.
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Silver chloride produced : = 46.149 g
Limiting reagent : CuCl2
Excess remains := 3.74 g
<h3>Further explanation</h3>
Reaction
silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate
Required
silver chloride produced
limiting reagent
excess remains
Solution
Balanced equation
2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)
mol AgNO3 :
= 58.5 : 169,87 g/mol
= 0.344
mol CuCl2 :
=21.7 : 134,45 g/mol
= 0.161
mol ratio : coefficient of AgNO3 : CuCl2 :
= 0.344/2 : 0.161/1
= 0.172 : 0.161
CuCl2 as a limiting reagent
mol AgCl :
= 2/1 x 0.161
= 0.322
Mass AgCl :
= 0.322 x 143,32 g/mol
= 46.149 g
mol remains(unreacted) for AgNO3 :
= 0.344-(2/1 x 0.161)
= 0.022
mass AgNO3 remains :
= 0.022 x 169,87 g/mol
= 3.74 g
