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Sergeeva-Olga [200]
3 years ago
8

How to balance an ionic equation BaCl2 + 2NH4CH3COO

Chemistry
1 answer:
Lorico [155]3 years ago
4 0
Assuming that the reaction proceeds, the reaction would be a double displacement reaction. BaCl2 + 2NH4CH3COO -> Ba(CH3COO)2 + 2NH4Cl If the reaction occurs in an aqueous solution, the products are soluble in water, so the reaction is probably not that interesting to look at.
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During the summer after your first year at Carnegie Mellon, you are lucky enough to get a job making coffee at Starbucks, but yo
Studentka2010 [4]

Answer:

The answer is "36.76  ml "

Explanation:

In the given question, Its basic understanding to also be hired seems to be that coffee heat = heat that the milk absorbs  

The following relationship defines heat:

\to q = m C_{p}( T_2 - T_1)

Its updated equation is:

\to q_{\ coffee} = - q_\ milk} \\\\ \to m_{\ milk} C_{p} (T_2 - T_1)_{\ coffee} = - m_{ \ milk} C_{p} (T_2 - T_1)_{\ milk} ..........(a)

The following information is provided to you as per the question:  

1) Milk and coffee have the same heat power  

2) Difference in coffee temperature is:

\to (T_2 - T_1)_{\ coffee} = 75.0^{\circ} \ \ C-85.0^{\circ} \ \ C

                              = -10.0 ^{\circ} \ \ C

3) The difference in milk temperature:

\to (T_2 - T_1)_{ \ milk} = 75.0^{\circ} \ \ C- 7.0^{\circ} \ \ C

= 68.0 ^{\circ} \ \ C

4)  Milk and coffee density are equal therefore, weight and volume are equal and have same ratio, that can be shown as follows:

\to \ Density (\rho) =\frac{ \ Mass (m)}{ \ Volume (V)} \\\\\to m= V \times  \rho \\\\

\to m_{\ coffee} \times (T_2 - T_1)_{ \ coffee}= -m_{\ milk} \times (T_2 - T_1)_{milk} \\\\\to V_{\ coffee} \times  \rho_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times  \rho_{\ milk} \times (T_2 - T_1)_{\ milk} \\\\\to V_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times (T_2 - T_1)_{\ milk}.............(b)

by replace the equation values (b) for calculating milk volumes as shown as follows:

\to V_{\ coffee} \times (T_2 - T_1)_{\ coffee} = - V_{\ milk} \times (T_2 - T_1)_{\ milk}\\\\\to 250 ml \times (-10 .0^{\circ} \ \ C) =  - V_{\ milk} \times (68.0^{\circ} \ \ C)\\\\\to V_{\ milk}= \frac{250 ml \times (-10 .0^{\circ} \ \ C) }{- (68.0^{\circ} \ \ C)}

               = 36.76 \ ml

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