(n + 11)(n - 4)
The answer is 1,83 moles.
The formula for the synthesis of ammonia from diatomic nitrogen and hydrogen is: N2+3H2-->2NH3
1 mole N2 = 28.0134g 1 mole N2 = 6.022 x 1023 molecules N2 28.0134g N2 = 6.022 x 1023 molecules N2 (4.00 x 1023 molecules N2) x (28.0134g/6.022 x 1023 molecules) = 18.6g N2
NO2 or N2
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
n2 + 7n - 44 (n + 11)(n - 4)
To factor a trinomial (three-term expression) of the form n2+bn+c, find the two numbers h and k that are factors of c and add up to b. Then, write those numbers in this template: (n+h)(n+k) For the trinomial n2+7n-44, c = -44 and b = 7. The two numbers that are factors of -44 and add up to 7 are 11 and -4. So, the factored form would be (n+11)(n-4).
38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...38.The position to value rule is Un= (n2+ 7n - 2)/2 where n = 1, 2, 3, ...
m(12 + 7n)
-7n2 + 74n + 33 = -7n2 + 77n - 3n + 33 = -7n*(n - 11) - 3*(n - 11) = =(-7n - 3)*(n - 11) = -(7n +3)*(n - 11)
100m^2-49n^2 (10m+7n)(10m-7n) the middle term cancels out.
-mn(m - n2)(m + n2)
.04m2-n2=(.2m+n)(.2m-n)
(7n - 4)(n + 3)
n2 + 9n + 18 = (n + 6)(n + 3)
No but it is composite factor of 44
Use the formula (n2 + 7) - n2 = 7, expand the equation to get n2 + 7n + 7n + 49 - n2 = 7, eliminate what you can to get 14n + 49 = 7, then isolate 'n' by subtracting 49 from both sides of the equation which will now be 14n = -42, now you divide both sides of the equation by 14 to get n = -3. Since 'n' is squared, you know that the two possible combinations of numbers can be either positive or negative. Their equation will look like this; 16 - 9 = 7, or 42 - 32 = 7, or (-42) - (-32) = 7.