Answer:
4.) 9, 1, and 4 5.) 4, 1, and 4
Explanation:
I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:
number 4
H: 9
P: 1
O: 4
number 5:
H: 4
S: 1
O: 4
you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.
hope this helped you :)
Answer: the second option: <span>Iron is being oxidized
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Explanation:
1) Oxidation is the increase of the oxidation state (number) due to the loss of electrons.
2) In the given reaction, you can see that in the left side the atom is Fe.
When an element (atom) is not combined (or combined with it self) its oxidation state is 0.
3) In the right side of the given equation you that iron is now in form of cation with charge 2+: Fe²⁺.
That means that the new oxidation state of the element is 2+.
4) This change in the oxidation state, of course, is accompanied by the loss of the two electrons: 2e⁻.
5) Conclusion: the iron has oxidized by losing two electrons and increasing its oxidation state from 0 to 2+.
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Answer:
V₂ = 2.91 L
Explanation:
Given data:
Initial volume = 3.50 L
Initial temperature = 90.0°C (90+273 = 363 K)
Final temperature = 30.0 °C ( 30 +273 = 303 K)
Final volume = ?
Solution:
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
V₁/T₁ = V₂/T₂
3.50 L / 363 K) = V₂ / 303 K)
V₂ = 0.0096 L/K × 303 K
V₂ = 2.91 L
Answer:
1040
Explanation:
you can use the mole to mole ratios of aluminum chloride and hydrogen gas,you first have to find the number of moles of hydrogen gas..the number of moles of hydrogen gas are 11.7(23.4/2)
2AlCl3+3H2
2 : 3
x :11.7
3x/3=23.4/3
x=7.8g/mol
then find the number of grams of aluminum chloride
m=n×mr
=7.8×133.341
=1040g
I hope this helps and sorry if it's wrong
