The moles of silver nitrate required to form 0.854 mol silver ion has been 0.854 mol.
The balanced chemical equation for the dissolution of silver nitrate has been:
The aqueous solution of silver nitrate has been dissociated into the constituent silver and nitrate ions.
In a balanced chemical equation, the coefficient has been equivalent to the moles of each reactant forming the moles of product.
From, the balanced equation,
The given moles of Silver has been 0.854 mol. Thus, the moles of silver nitrate required has been given as:
The moles of silver nitrate required to form 0.854 mol silver ion has been 0.854 mol.
For more information about the moles produced, refer to the link:
brainly.com/question/10606802
Answer:
b- swap the uses for iron oxide and calcium carbonate.
Explanation:
just took the test. good luck!
The mass is 1610 g.
Explanation:
<span>D=<span>MV</span></span>, where D is density, M is mass, V is volume
Known/Given
D=0.921 g/mL
<span>V=1.75 L×<span><span>1000mL</span><span>1L</span></span>=1750 mL</span>
Unknown
M
Equation
<span>D=<span>MV</span></span>
Solution
Rearrange the equation to isolate the mass and solve for mass.
<span>M=D×V=0.921g/mL×1750mL=1611.75 g=1610 g</span> with three significant figures.
The following diagram helps to understand how to manipulate the density equation so solve for any of the variables.
The following Density Triangle will help you determine how to solve
for any missing variable. Put your finger over the missing variable to
see what you need to do to solve for it. For example, if you need to
find volume, put your finger over the V, and you get <span>M÷D</span>.
https://www.teacherspayteachers.com/Product/Science-Journal-Density-Triangle-Foldable-Completed-4360...
<span>[Ag+]3[PO4]3-
Ag3PO4(aq) ----> 3Ag^+(aq) + PO4^3-(aq)
In pure water, the molarity of PO4^3-, which is 1.76x10^-5 M, also represents the molar solubility of Ag3PO4, that means that max 1.76x10^-5 mol Ag3PO4 dissolves in 1L (1000 mL) of water. When the amount of water is halved to 500 mL, the solubility of Ag3PO4 is also halved. (8.8x10^-5 mol).
Ag3PO4(aq) ----> 3Ag^+(aq) + PO4^3-(aq)
8.8x10^-5 mol ..... 3x8.8x10^-5 mol = 2.64x10^-4 mol
[Ag^+] = n / V = 2.64x10^-4 mol / 0.500 L = 5.28x10^-4 M</span>
Explanation:
35- this is the answer in 2s.f
