Answer:
in position and time graphs the slope of line indicates the velocity
Answer:
The option<u><em> D) increases reproduction of native species</em></u> is NOT an impact that an invasive non-native species has on an ecosystem.
Explanation:
Introduction of a non-native species into a stable ecosystem can be very threatful to the ecosystem. The non-native species will compete for resources with the native species of the area. As a result, there will be severe competition. The competition might lead to migration of the native species or them being endangered on the whole. Also, the introduction of a non-native species will result in the extinction of a particular prey as it will have an increase in the number of predators.
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Explanation:
Unbalanced reaction equation:
H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O
We can use a simple mathematical approach to balance this equation by solving simple algebraic equations.
aH₂SO₄ + bNH₄OH → c(NH₄)₂SO₄ + dH₂O
a,b,c and d are simple coefficients that will balance the equation:
Conservation of H: 2a + 5b = 8c + 2d
S: a = c
O: 4a + b = 4c + d
N: b = 2c
lets assume that a = 1
c= 1
b = 2
Solving for d from 4a + b = 4c + d
d = 4a + b -4c = 4(1) + 2 - 4(1) = 2
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Learn more:
Balancing of equations brainly.com/question/6078553
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Answer:
The answer is 139.1 g of sodium azide.
Explanation:
To solve this question we need to find first the stoichometry of the reaction to find the moles involved.
1. The balanced formula:
2NaN3 ⇒ 3N2 + 2Na
(important: the nitrogen is a gas, therefore, is always diatomic N2)
2. The moles:
We need to convert the 90 g of N2 into moles to know the relations of amounts between the other compounds in the chemical equation
90 g of N2 x (1 mol of N2/ 28 g of N2) = 3.21 mol of N2
3. Stoichometry:
The relations of the amounts of moles of N2 and NaN3 using the balanced formula
3.21 mol of N2 x (2 mol of NaN3/3 mol of N2) = 2.14 mol of NaN3
4. Moles into grams
then to answer the question we convert the moles value of NaN3 into grams of NaN3
2.14 mol of NaN3 x ( 65 g of NaN3/ 1 mol of NaN3) = 139.1 g of NaN3.
and this is the answer of the question.
