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LenKa [72]
3 years ago
15

Explain wether or not a reaction takes place when when bromine water is added to sodium chloride solution

Chemistry
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

When chlorine (as a gas or dissolved in water) is added to sodium bromide solution, the chlorine takes the place of the bromine. Because chlorine is more reactive than bromine, it displaces bromine from sodium bromide. The solution turns brown. ... The chlorine has gone to form sodium chloride.

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Is the indicator generally added to the titrant or the analyte in a titration?
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In a titration process, the unknown or the analyte with a known volume is placed in a flask and the titrant whose concentration is known is placed in the burette. The indicator in the titration process is generally added to the flask with the analyte. 
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The reaction between alcohols and carboxylic acids is called
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Explanation:

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4 0
3 years ago
A system gains 687 kJ of heat, resulting in a change in internal energy of the system equal to 156 kJ. How much work is done?
Maslowich

Answer:

w = -531 kJ

1. Work was done by the system.

Explanation:

Step 1: Given data

  • Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
  • Change in the internal energy of the system (ΔU°): 156 kJ

Step 2: Calculate the work done (w)

We will use the following expression.

ΔU° = q + w

w = ΔU° - q

w = 156 kJ - 687 kJ

w = -531 kJ

By convention, when w < 0, work is done by the system on the surroundings.

4 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
4 years ago
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