261.162 grams. Use the equation n=M÷Mr.
Answer:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:

I believe that it is A. If you remember in lesson 03.03 it gave multiple examples of things that support the continental drift theory. One of them was fossils of the same organisms found in different continents.
The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation.
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
Answer:
Lichens can be used as air pollution indicators, especially of the concentration of sulphur dioxide in the atmosphere. Lichens are organisms that grow in exposed places such as rocks or tree bark. They need to be very efficient at absorbing water and nutrients to grow there.