You did not include the options but I can tell you the product ratio.
The product ratio is the mole ratio of the products of the reaction.
From the balanced chemical equation you have all the mole ratios:
The given equation is: 2 C6H5COOH + 15O2 --> 14 CO2 + 6H2O
The mole ratios are: 2 C6H5COOH: 15 O2: 14 CO2 : 6 H2O
The products are CO2 and H2O
Their mole ratio = 14 CO2 : 6 H2O
That can be expressed as:
14 mol CO2 7 mol CO2
----------------- = -----------------
6 mol H2O 3 mol H2O
It is also the same that:
6 mol H2O : 14 mol CO2
6 mol H2O 3 mol H2O
------------------ = -------------------
14 mol CO2 7 mol CO2
So, compare your options to the ratios show above and pick the proper ratio.
Answer:
1.315x10⁻³M = [Ca²⁺]
Explanation:
Based in the reaction:
Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)
Solubility product, ksp, is defined as:
ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²
From 1 mole of hydroxyapatite are produced 10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:
6/10 Ca²⁺ = PO₄³⁻
Replacing in ksp formula:
ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²
As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:
2.34x10⁻⁵⁹ = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²
3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶
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<em>1.315x10⁻³M = [Ca²⁺]</em>
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I hope it helps!
Answer:
Yes
Explanation:
1. Mass of 0.60 mol of AuCl₃
2. Mass of AuCl₃ in 750 mL
The solubility of AuCl₃ is 68 g/100 mL.
In 750 mL of water, you can dissolve
∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.
given E = 9.4145E-25
h = 6.626E-34
c = 2.998E8
sub values into the equation above, and solve for wavelength.
You will get 0.211m
Remember: heat lost = heat gained
When calculating heat loss or gain, remember
mass*(spec heat cap)*(change in T)
The unknown loses heat- we don't know the spec heat cap, so we'll call it x.
The water gains. I've omitted the units, but always use when solving problems on your own.
75*x*(96.5-37.1) = 1150*4.184*(37.1-25)
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Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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