All animals are _______. a. producers b. omnivores c. herbivores d. consumers Please select the best answer from the choices pro
2 answers:
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Here we have to calculate the amount of ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
Hi, P.S. The atomic number will always be the same as the number of protons...
Answer:
The energy released will be -94.56 kJ or -94.6 kJ.
Explanation:
The molar mass of methane is 16g/mol
The given reaction is:
the enthalpy of reaction is given as ΔH = -890.0 kJ
This means that when one mole of methane undergoes combustion it gives this much of energy.
Now as given that the amount of methane combusted = 1.70g
The energy released will be: