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Determine how many mL of 0.102M NaOH solution are needed to neutralize 35.0 mL of 0.125M H2SO4 solution. Show the given, balanced equation, solution and final answer.
2NaOH+H2SO4→2H2O+2Na+SO4moles of sulfuric acid = (0.035 L)(0.125 M) =0.0044mol0.0044× 2× 0.102= 0.0008976mol39.997 × 0.0008976= 0.036
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Given the balanced equation: H2SO4 + 2NaOH ----- 2H2O + Na2SO4.What volume of a 1.450 M NaOH solution is required to titrate 35.00 mL of a 4.50 M H2SO4 solution? A.158 mL B.163 ml C.217 ml D.236 ml
C.217 ml
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A 100 mL sample of a 6.0 mol/L solution of H2SO4 is diluted to a final volume of 500 mL. What is the concentration of the diluted solution?
For answering this question you need to calculate amount of mol in final mixture. 6.0 mol in 1 L x mol in 0.1L x = 0.6 mol So you have...
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6. Calculate what volume of a sulfuric acid solution (specific gravity = 1.07; and containing 10.0 % H2SO4 by mass) contains 18.50 g of pure H2SO4 at a temperature of 25°C.
1.07 g/mL x ?# mL x 0.10 = 18.50 g.0.107ml=18.50g= 172.9ml
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Calculate the work necessary to compress reversibly 0.200 mol of an ideal gas at 300 K from an initial volume of 4.00 L to a final volume of 2.00 L
Work required to compress=+1718.3J
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calculate normality and titre of koh -solution (volume 5.45 ml) if it is necessary for the titration 6.7 ml h2so4-solution with cn=0.1025 mol/l
Solution:CN(KOH) × V(KOH) = CN(H2SO4) × V(H2SO4)Therefore,CN(KOH) = CN(H2SO4) × V(H2SO4) / V(KOH) CN(KOH) = (0.1025 mol L-1 × 6.7 mL) / (5.45 mL) = 0.1260 mol L-1Normality of KOH...
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To what volume should you dilute 90 mL of a 11.5 M H2SO4 solution to obtain a 1.65 M H2SO4 solution?
Dilution ratio: 11.5 / 1.65 = 6.97 Target volume: 90 ∙ 6.97 = 627.3 mL
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Using 10 ml of 12 M HCl stock solution, calculate the volume of water necessary to
create a 0.01 M solution of HCl. What is the final volume of the solution?
Given:V1=10mL, C1=12 M, C2=0.01 MFind: V2 - ?Solution:Since the quantity of HCl remains the same after dillution C1xV1=C2V2V2= (C1xV1)/V2=(12 M x 10mL)/0.01 M = 12 x 103 mLAnswer: The final...
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Using 2 ml of 19 M NaOH, calculate the volume of water necessary to create a 0.25 M
solution of NaOH. What is the final volume of the solution
C1(NaOH) = 19 M;C2(NaOH) = 0.25 M;V1 = 2 ml;V2 = unknown.Proportion 1:19 moles — 1000 mlx moles — 2 mlx = 0.038 moles in 2 ml of 19 M...
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a 25ml sample of dilute h2so4 is titrated with 0.0965m naoh so that both protons of h2so4 are neutralized it takes 17.83ml of base to reach end point what is the molarity of h2so4
Given,Volume of H2SO4, (V1) = 25mlStrength of NaOH = 0.0965 m For the titration purpose Normal strength of NaOH will usedSo,Normally of the NaOH = (0.0965×no. of OH– in 1...
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