345238+11000+5000+1200+1000+1500+2000+35+30+216330+8035306 =8618639 All the operations are addition so this can all be added in one step.
These numbers all add up to 81.
300000000+9000000+200000+70000+30+1 =309270031 All the addition can happen in the same step. When all the operations are the same it can be done that way.
813+813+812+808+806+805+804+804+804+803+802+796+792+792+783 =12037 All the operations are addition so all the adding can happen in one step.
All six of them
When they react K2SO4 and BaBr2 they'll give 2KBr and BaSO4: SO42-aq + Ba2+aq ---> (BaSO4)s ('s' = solid = precipitating bariumsulfate) 2K+ and Ba2+ are spectator ions (all 'aq' = hydrated in solution)
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Solubility is normally measured as the amount of a substance that can dissolve in a solvent and is measured in units of (amount of substance)/(volume of solvent).Barium nitrate, for example, has a solubility of 0.105 moles/litre in water, or it can also be expressed as 27.3g/litre.The solubility product is an equilibrium constant for the process:salt (solid) cations (aq) + ions (aq)As the activity of the solid is taken to have a value of 1, the solubility product is the result of multiplying the concentrations of the dissolved ions at the solid's maximum solubility.The solubility product (Ksp) can be thought of as a more general expression of the solubility of a substance. To obtain the solubility product, we multiply the concentrations (mol/litre)* of ALL the ions that result from the substance dissolving when they are at their maximum concentration.In the case of barium nitrate, we have:Ba(NO3)2 ---> Ba2+ + NO3- + NO3-So the solubility product , Ksp = [Ba2+][NO3-][NO3-] = [Ba2+][NO3-]2The experimentally determined value of the solubility product for Barium Nitrate is 4.64×10-3.For barium nitrate dissolving in pure water, [NO3-]= 2[Ba2+], so we can writeKsp = [Ba2+][NO3-]2 = 4[Ba2+]3 = 4.64×10-3Solving this, [Ba2+] = 0.105 mol/litreAs [Ba2+] equals the concentration of barium nitrate that is dissolved, this is the solubility of barium nitrate in pure water.The usefulness of the solubility product is that we can use it to look at the effects of ions from other sources than the salt. For example, if we try to dissolve barium nitrate in 10 molar nitric acid, we use the same expression as before:Ksp = [Ba2+][NO3-]2 = 4.64×10-3But this time [NO3-] = 10 mol/litre (approx), as that is already in solution, so:100 [Ba2+] = 4.64×10-3[Ba2+] = 4.64×10-5The solubility of barium nitrate in 10M nitric acid is 4.64×10-5 Mol/litre, MUCH less than its solubility in pure water, and not something that you could determine from the solubility alone.*Strictly, we should use activity rather than concentration.
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At saturated solution of (partly) insoluble salt (imaginable: A2B) the following Equilibrium exists:(A2B)s --> 2A+ + B2-The thermodynamic Equilibrium constant Ksol equals: Ksol = [A+]2*[B2-] , the square hooks symbolise the ion concentration in mole/L. Note also the power in superscript [A]power which is equal to the number of ions in the (standardised) reaction equation. Further it should be noted that the amount of solid salt (A2B)s is not in the nominator of this Equilibrium equation, Ksol = [A+]2*[B2-] and that's why it is called (Equilibrium) solubility product.
Barium is a group 2 element with a 2+ charge, making it iso-electronic with ions such as calcium (Ca2+), strontium (Sr2+), and radium (Ra2+). These ions have the same electronic configuration as the barium ion.
8+8+8+6+6+16 =52 All the addition can happen in the same step. When the operations are all the same you can do this.
1+12+123+1234+12345+123456+1234567+12345678+123456789+12345678910+1234567891011+123456789101112 =124703839845238 All the operations are addition so all the addition can happen in one step.
345238+11000+5000+1200+1000+1500+2000+35+30+216330+8035306 =8618639 All the operations are addition so this can all be added in one step.
7000040566 Yep that it all right :|
If you do not assume that "plus" is commutative, all that can be said that it is equal to A plus B.
If you add all the numbers from 1 to 98, the total is 4851.