If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.
<h3>What is a strong acid?</h3>
A weak acid is one that is able to ionize completely in solution. The acid called chromic acid H2CrO4 is not able to ionize completely in solution.
We know that a buffer is composed of a weak acid and its salt or a weak base and its salt hence if the acid H2CrO4 is present in a buffer then the other half must be salt of the acid.
If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.
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Answer:
Explanation:
Incorrect use has resulted in accidents and disfiguring burns. Never leave a methylated spirit appliance unattended. Also make sure that the camping stove or appliance is on a flat surface and that the fuel cannot spill out.
Answer:
The answer to your question is: letter A
Explanation:
A combination reaction is when there are two reactants that gives only one product.
a. 2SO2 + O2—> 2SO3 This is a combination reaction,
2 reactants gives one product.
b. Zn + Cu(NO3)2–>Zn(NO3)2 + Cu This is not a combination reaction,
it's a single replacement reaction.
c. 2H2O2–> 2H2O+O2 This is a decomposition reaction
d. AgNO3 + NaCl → AgCl+NaNO3 THis is a double replacement reaction.
Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.
