Answer: 1.98 g
Explanation:
To calculate the moles :
The balanced given equation is:
According to stoichiometry :
4 moles of
will produce = 2 moles of
Thus 0.22 moles of
will produce=
of
Mass of
Thus 1.98 g of water is produced along with 5.0 L of
at STP
Answer:
d. Pergoldic acid
Explanation:
chlorate = ClO₃⁻
Goldate GoO₃⁻
On this basis we can infer that HClO₄ and HGoO₄ will be similar .
HClO₄ is called perchloric acid , hence HGoO₄ will be called pergoldic acid.
Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
6NaC₂H₃O₂ + Fe₂O₃ → 2Fe(C₂H₃O₂)₃ + 3Na₂O
Explanation:
Given equation;
NaC₂H₃O₂ + Fe₂O₃ → Fe(C₂H₃O₂)₃ + Na₂O
To find the coefficient that will balance this we equation, let us set up simple mathematical algebraic expressions that we can readily solve.
Let us have at the back of our mind that, in every chemical reaction, the number of atom is usually conserved.
aNaC₂H₃O₂ + bFe₂O₃ → cFe(C₂H₃O₂)₃ + dNa₂O
a, b, c and d are the coefficients that will balance the equation.
conserving Na; a = 2d
C: 2a = 6c
H: 3a = 9c
O; 2a + 3b = 6c + d
Fe: 2b = c
let a = 1
solving:
2a = 6c
2(1) = 6c
c = 
2b = c
b =
= 
d = 2a + 3b - 6c = 2(1 ) + (3 x
) - (6 x
) = 
Now multiply through by 6
a = 6, b = 1, c = 2 and d = 3
6NaC₂H₃O₂ + Fe₂O₃ → 2Fe(C₂H₃O₂)₃ + 3Na₂O
learn more:
Balanced equation brainly.com/question/9325293
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Answer:
The importance of significant figures
As stated before, it is important within the science fields that you are not more precise or accurate than the least accurate or precise number. In science, it is generally agreed upon that the last number digit in any figure is filled with uncertainty.
Explanation: