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A barium hydroxide,Ba(OH)2 solution reacts with a nitric acid solution, HNO3 according to the following balanced equation: Ba(OH)2+2HNO3=Ba(NO3)+2H2O Calculate the amount of energy released during the reaction if 0,18 moles of Ba(OH)2 reacts completely with the acid

A barium hydroxide,Ba(OH)2 solution reacts with a nitric acid solution, HNO3 according to the following balanced equation: Ba(OH)2+2HNO3=Ba(NO3)+2H2O Calculate the amount of energy released during the reaction if 0,18 moles of Ba(OH)2 reacts completely with the acid

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Hy guys i need a physics and chemistry tutor and I'm going to rewrite so i need all the help I can get

Hy guys i need a physics and chemistry tutor and I'm going to rewrite so i need all the help I can get

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Guys I need help on physics.... Drop your WhatsApp number if you're interested NB:Girls only

Guys I need help on physics.... Drop your WhatsApp number if you're interested NB:Girls only

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Sorri but none of the video's is playing...does anyone else have a problem with this or is it just me

Sorri but none of the video's is playing...does anyone else have a problem with this or is it just me

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I tried to watch videos and they responde by " This is private video" help plz

I tried to watch videos and they responde by " This is private video" help plz

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The circular motion of a particle in the x-y is given by the equations: r(t)= 7.23m Theta(t)= (6.33rad/s)t Where r is the distance from the origin and Theta is the polar angle measure counterclockwise from the positive x-axis. Calculate the x-coordinate of the particle at time t=2.50s

The circular motion of a particle in the x-y is given by the equations: r(t)= 7.23m Theta(t)= (6.33rad/s)t Where r is the distance from the origin and Theta is the polar angle measure counterclockwise from the positive x-axis. Calculate the x-coordinate of the particle at time t=2.50s

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Are you guys going to organize tutorial workshops for Physics as well? They are extremely helpful for Maths.

Are you guys going to organize tutorial workshops for Physics as well? They are extremely helpful for Maths.

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an athlete runs 100m in 10.0 seconds. assume that he accelerates uniformly from rest for the 20m and then he runs at constant velocity for the remainder of the race. find the final velocity and the time taken accelerating

an athlete runs 100m in 10.0 seconds. assume that he accelerates uniformly from rest for the 20m and then he runs at constant velocity for the remainder of the race. find the final velocity and the time taken accelerating

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When a 0.998g sample of CuSO4.xH20 was heated strongly,so that the waters of hydration were given off,the mass of the anhydrous salt remaining was found to be 0.638 g.What is the experimental value of the percent water of hydration?

When a 0.998g sample of CuSO4.xH20 was heated strongly,so that the waters of hydration were given off,the mass of the anhydrous salt remaining was found to be 0.638 g.What is the experimental value of the percent water of hydration?

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Subject: Heat of mixingnnnHi Therenn[Sigurd Skogestad, page 113, question 4.7b].I ve tried this question at it is on the textbook,but i get a different answer,where am I going wrong? nn For the following assume the same initial temperatures before mixing.(20Â°C) a. Calculate the adiabatic temperature rise when we mix 0.5kg sulphuric acid with 1kg water. nYou are given the following information: nHeat of mixing = -96kJ/mol H2SO4nMolecular weight for the acid = 98.1g/mol Specific heat capacity of water= 4.18 kJ/kg K nnnnn Solutionsna. 500g =500/98.1 = 5.09 mol H2SO4 nn1kg = 1000/18 = 55.5 mol H2OnEnthalpy mix = -96*5.09 = -489.3 kJ/kg solution (used -96kJ/mol because mol of water >10) Temperature change = -489.3/4.18=117 K.nnTextbook answer is 72 K

Subject: Heat of mixingnnnHi Therenn[Sigurd Skogestad, page 113, question 4.7b].I ve tried this question at it is on the textbook,but i get a different answer,where am I going wrong? nn For the following assume the same initial temperatures before mixing.(20Â°C) a. Calculate the adiabatic temperature rise when we mix 0.5kg sulphuric acid with 1kg water. nYou are given the following information: nHeat of mixing = -96kJ/mol H2SO4nMolecular weight for the acid = 98.1g/mol Specific heat capacity of water= 4.18 kJ/kg K nnnnn Solutionsna. 500g =500/98.1 = 5.09 mol H2SO4 nn1kg = 1000/18 = 55.5 mol H2OnEnthalpy mix = -96*5.09 = -489.3 kJ/kg solution (used -96kJ/mol because mol of water >10) Temperature change = -489.3/4.18=117 K.nnTextbook answer is 72 K

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