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Irina-Kira [14]
4 years ago
6

Consider the following double replacement reaction between silver nitrate and sodium chloride: AgNO_3(aq) + NaCl(aq) → AgCl(s) +

NaNO_3(aq) \DeltaH^{\circ}_{rxn} Which of the following is the correct expression needed to calculate\DeltaH^{\circ}_{rxn}?
Chemistry
1 answer:
Rama09 [41]4 years ago
5 0

The question is not complete. Proposed completed question is:

When aqueous solutions of sodium chloride and silver nitrate are mixed, a double-displacement reaction occurs.  What is the balanced equation for the reaction?

               A.  

NaCl(aq) + AgNO3(aq) --------NaNO3(aq) + AgCl(aq)

               B.  

NaCl(aq) + AgNO3(aq) -------- NaNO3(aq) + 3AgCl(s)

               C.  

NaCl(aq) + AgNO3(aq) --------- NaNO3(aq) + AgCl(s)

               D.  

NaCl(aq) + AgNO3(aq) --------- Na(s) + AgNO3Cl(aq)

               E.  

NaCl(aq) + AgNO3(aq) --------- NaNO3(aq) + ClAg(s)

Answer is

       A.  

NaCl(aq) + AgNO3(aq) --------NaNO3(aq) + AgCl(aq)

Explanation:

Following the law of conservation of energy, the total number of atoms in the reactant  side is equal to the total number of atoms in the product side.

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`Determining the limiting reactant is an extremely important concept in chemistry. To do this, you need to know all of the follo
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Read 2 more answers
A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23
Anna35 [415]

Answer:

0.8696\ \text{J/g}^{\circ}\text{C}

Explanation:

m_m = Mass of metal = 19 g

c_m = Specific heat of the metal

\Delta T_m = Temperature difference of the metal = 99-23=76^{\circ}\text{C}

V = Volume of water = 150 mL = 150\ \text{cm}^3

\rho = Density of water = 1\ \text{g/cm}^3

c_w = Specific heat of the water = 4.186 J/g°C

\Delta T_w = Temperature difference of the water = 23-21=2^{\circ}\text{C}

Mass of water

m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}

Heat lost will be equal to the heat gained so we get

m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}

The specific heat of the metal is 0.8696\ \text{J/g}^{\circ}\text{C}.

4 0
3 years ago
Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro
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Answer:  8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation:  Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3.  1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

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We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm)  Round to 3 sig figs (856 cm and 8560 mm)

5 0
3 years ago
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