Metamorphic rocks, because sedimentary rocks are formed from compressed sand, gravel, dirt, etc. and igneous rocks are formed by cooled magma, but of course each type of rock can be formed other ways. so it leaves us with metamorphic.
2Ag +(aq) + SO4 2-(aq)----->Ag2SO4 (s)
Answer:
The answer to your question is below
Explanation:
a) CI Chlorine is located in group VIIA, so its must gain one electron to be stable.
b) Se Selenium is located in group VIA, it must gain 2 electrons to be stable.
c) N Nitrogen is located in group VA, it must gain 3 electrons to be stable.
d) I Iodine is located in group VIIA, so it must gain 1 electron to be stable.
e) S sulfur is located in group VIA, so it must gain 2 electrons to be stable.
Answer:
0.005 M
Explanation:
Given data:
volume of sample solution ( volume of D ) = 5.0 mL
volume of added stock solution ( V1 ) = 5.0 mL
concentration of added stock solution ( N1 ) = 0.02 M
Total volume of concentration ( V2 )= 10 mL = ( 5.0 mL + 5.0mL)
concentration of Total volume of sample ( C2 ) = 0.01
N2 = ( N1V1 ) / V2
= ( 0.02 * 5 ) / 10 = 0.01 m
absorbance of sample solution ( A1 ) = 0.10
absorbance of additional sample solution ( A2 ) = 0.20
attached below is the remaining part of the detailed solution
Answer:
H₂O.
Explanation:
- It is clear from the balanced equation:
<em>CH₄ + 2H₂O → CO₂ + 4H₂.</em>
that 1.0 mole of CH₄ reacts with 2.0 moles of H₂O to produce 1.0 mole of CO₂ and 4.0 moles of H₂.
- To determine the limiting reactant, we should calculate the no. of moles of (20 g) CH₄ and (15 g) H₂O using the relation:
<em>n = mass/molar mass</em>
<em></em>
no. of moles of CH₄ = mass/molar mass = (20 g)/(16 g/mol) = 1.25 mol.
no. of moles of H₂O = mass/molar mass = (15 g)/(18 g/mol) = 0.833 mol.
- <em>from the balanced reaction, 1.0 mole of CH₄ reacts with 2.0 moles of H₂O.</em>
So, from the calculated no. of moles: 0.4167 mole of CH₄ reacts completely with 0.833 mole of H₂O and the remaining of CH₄ will be in excess.
<u><em>So, the limiting reactant is H₂O.</em></u>
