Answer:
1.10 g H2
Explanation:
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Answer:

Explanation:
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In this case, since this imaginary gas can be modelled as an ideal gas, we can write:

Which can be written in terms of density and molar mass as shown below:

Thus, by computing the pressure in atmospheres, the resulting density would be:

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B is the correct answer since both Gondwanaland and Laurasia both technically came after pangaea
Answer: 0.4
Explanation: The easiest way to do this is to get a denominator of 10. To do this, multiply each number in 2/5 by 2. You will get the fraction 4/10. With the denominator of 10, the answer would be 0.4 by putting the decimal in front of the number.
This is the three cases that help to determine the minimum concentration of KOH required for precipitation
Part a) 1.5×10^−2 M K CaCl2
Part b) 2.3×10^−3 M Fe (NO3)2
Part c) 2.0×10^−3 M MgBr2
a) CaCl2 + 2KOH --> Ca (OH) 2 + 2KCl Ca (OH) 2 <=> Ca^2+ + 2OH^-
ksp = 1.5*10^-2 + x^2
4.68*10^-6 = 1.5*10^-2 + x^2
x= [KOH] = 0.01766
b) Fe (NO3)2 +2 KOH--> Fe (OH)2 + 2KNO3
Fe (OH)2 <=> Fe^2+ + 2OH^-
ksp = 2.3*10^-3 + x^2
4.87*10^-17 = 2.3*10^-3 + x^2
x= 1.46*10^-7
c) MgBr2 + KOH --> Mg (OH) 2 + 2KBr
Mg (OH) 2 <=> Mg^2+ + 2OH^-
ksp = 2.0*10^-3 + x^2
2.06*10^-13 = 2.0*10^-3 + x^2
x= 1.015*10^-5