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Olin [163]
3 years ago
9

A chemical reaction produces 56.2 grams of iron (II) sulfide (FeS). How many moles of iron (II) sulfide does the reaction produc

e? (1 mole FeS = 87.92 grams FeS) *
1.00 moles FeS

2.50 moles FeS

0.639 moles FeS

4.98 moles FeS
Chemistry
1 answer:
Soloha48 [4]3 years ago
8 0

Answer:

0.639

Explanation:

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Answer:

                      Q  =  2647 J

Explanation:

                    Specific heat capacity is the amount of energy required by one Kg of a substance to raise its temperature by 1 °C.

In thermodynamics the equation used is as follow,

                                                 Q  =  m Cp ΔT

Where;

           Q  =  Heat  =  ?

           m  =  mass  =  660 g

           Cp  =  Specific Heat Capacity  =  0.3850 J.g⁻¹.°C⁻¹

           ΔT  =  Change in Temperature  =  23.35 °C - 12.93 °C  =  10.42 °C

Putting values in eq. 1,

                            Q  =  660 g × 0.3850 J.g⁻¹.°C⁻¹ ×  10.42 °C

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(select all that apply.) the elements most often found in organic molecules are _____.
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7 0
3 years ago
Read 2 more answers
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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