An isotope of Aluminium, a 'normal' Al atom would have a proton number of 13 (as this does not change in isotopes of the same element) but only 14 neutrons. This isotope of Al has 1 extra neutron than regular Al.
The molarity of the base ( ammonium hydroxide) is calculated as follows
find the moles of Nitric acid =molarity x volume
=29.5 x 0.175 =5.163 moles of nitric acid
write the equation for reaction
HNO3+ NH4OH =NH4NO3 +H2O
by use of mole ratio between HNO3 to NH4NO3 which is 1:1 the moles of NH4OH is also 5.163 moles
molarity =moles/volume
= 5.163 /50=0.103 M is the molarity of the base
Answer: The number of moles of excess reagent remain unreacted will be, 6.004 moles.
Explanation : Given,
Moles of 
= 1.332 mol
Mass of 
= 6.504 mol
First we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
From the balanced reaction we conclude that
As, 13 mole of react with 1 mole of
So, 6.504 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Number of moles remain unreacted = 6.504 mol - 0.5003 mol = 6.004 mol
Therefore, the number of moles of excess reagent remain unreacted will be, 6.004 moles.
Fishes: salmon, steelhead, rainbow trout, small minnow-like speckled dace, California roach and large minnow-like fishes.
Reptiles and amphibians: California red-legged frog, special-status northwestern pond turtles, Pacific tree frogs, Foothill yellow-legged frogs, <span>gopher and garter snake, alligator lizards and western fence lizards
Birds: California clapper rail and California black rail
Invertebrates: mayflies, caddisflies, stoneflies,water boatmen, water striders, water beetles, dragonflies, California oysters, native clams, red abalone, Dungeness crabs
Plants:sedges, rushes, pickleweed, cord grass, gum plant, eelgrass, and mosses</span><span>
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Answer:
V = 0.2714 mL
Explanation:
Henry's Law:
∴ Kh CO2 = 0.034 M/atm ....The Henry's constant
∴ Partial pressure: P CO2 = 3.7 atm
⇒ <em>C</em> CO2 = P CO2 / Kh
⇒ <em>C</em> CO2 = 3.7 atm / 0.034 M/atm = 108.823 M (mol/L)
∴ molar mass CO2 = 44.01 g/mol
⇒ mol CO2 = (1.3 E3 mg)(g/1000 mg)(mol/44.01 g) = 0.03 mol CO2
volumen solution:
⇒ V sln = (0.03 mol)(L/108.823 mol) = 2.714 E-4 L
⇒ V sln = 0.2714 mL
