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enot [183]
3 years ago
7

If an atom has a mass number of 25 and an atomic number of 11 how many neutrons are there in this atom?

Chemistry
1 answer:
babunello [35]3 years ago
5 0
C. 14
you would subtract the mass number from the protons or the atomic number because in order to find the mass number you would add the protons and neutrons :)
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What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?
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This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

M=\frac{mol\ solute}{Volume\ solution \ in \ L}

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L

Finally, we plug in these numbers in the equation for the calculation of molarity:

M=\frac{0.0313mol}{0.100L}\\\\M=0.313M

Learn more about molarity: brainly.com/question/10053901

8 0
2 years ago
When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
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