2.258625 *10²³ oxygen atoms will be produced.
<h3><u>Explanation:</u></h3>
Decomposition reaction is defined as the type of reaction where one single reactant breaks to produce more than one product only by means of heat or other external factor.
Formula of magnesium oxide = MgO.
The molecular mass of magnesium oxide = 24 +16= 40.
So in 40 grams of magnesium oxide, number of molecules is 6.023 * 10²³.
So in 15 grams of magnesium oxide,, number of molecules is 6.023 *1023 * 15/40 = 2.258625 *10²³.
From one molecule of magnesium oxide, one oxide atom will be produced.
So number of oxide atoms with 100% yeild = 2.258625 *10²³
The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
Answer:
300 mM
Explanation:
In order to solve this problem we need to calculate the line of best fit for those experimental values. The absorbance values go in the Y-axis while the concentration goes in the X-axis. We can calculate the linear fit using Microsoft Excel using the LINEST function (alternatively you can write the Y data in one column and X data in another one, then use that data to create a dispersion graph and finally add the line of best fit and its formula).
The <u>formula for the line of best fit for this set of data is</u>:
So now we <u>calculate the value of </u><u><em>x</em></u><u> when </u><u><em>y</em></u><u> is 1.50</u>:
Answer:
okey if I'm going on south America for the winter break i would basically pack baggy clothes ,sweater,jackets,wollen hats,wollen clothes,socks,shoes,scarf,heavy and wollen one thats all ig
Explanation:
When conducting a melting point experiment, if we were to heat a sample quickly. Large amount heat is provided instantly which would melt the crystals in the tube very quickly, even before the temperature of the thermometer reaches to that level. So the observes melting point would be much lower than the actual melting point when sample is heated slowly.