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enot [183]
3 years ago
5

Consider this reaction: NH + + HPO 4 + NH3 + H2PO4

Chemistry
1 answer:
Serggg [28]3 years ago
5 0

Answer:NH3

Explanation:Ecd

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They are called the terrestrial planets and are smaller than the others. They are he hottest planets and closest to the sun and take less time to rotate the sun.
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3 years ago
Question 22
ruslelena [56]

Answer:

O2, oxygen.

Explanation:

Hello.

In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

n_{CO_2}^{by\ CH_4}=85gCH_4*\frac{1molCH_4}{16gCH_4} *\frac{1molCO_2}{1molCH_4} =5.3molCO_2\\\\n_{CO_2}^{by\ O_2}=320gO_2*\frac{1molO_2}{32gO_2} *\frac{1molCO_2}{2molO_2} =5molCO_2

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

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5 0
3 years ago
How many grams of a 19.6% sugar solution contain 72.5 g of sugar?
anzhelika [568]

Answer:

6.2g of sugar solution contains 72.5g

4 0
3 years ago
Alll acids will eventually produce OH-.<br> True or False?
Ugo [173]
All acids contain OH in their formula
8 0
3 years ago
What is the percent composition of NaHCO3?
zhuklara [117]

Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

8 0
3 years ago
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