Bromine is used in many areas such as the following agricultural chemicals, dyestuffs, insecticides, pharmaceuticals and chemical intermediates. Some of the uses are being phased out for environmental reasons, but new uses can be found with Bromine can be used for furniture foam, plastic casings for electronics and textiles to make them less flammable
Answer:
Once ignited, a ribbon of magnesium metal burns freely in air to form solid magnesium oxide in the form of a white powder. The reaction is plainly spontaneous even though ΔS is negative.
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Now, as the stoichiometrical factors are in terms of mole but no information about neither the temperature nor the pressure is given, by means of the Avogadro's law, one could perform the stoichiometric calculations with the given volume as both the pressure and temperature remain the same, that is:
Such 1:1 volume relationship equals the 1:1 molar relationship given in the chemical reaction in terms of their stoichiometric coefficients, therefore, the yielded volume of carbon dioxide is also 9.23m³
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The equilibria showing how the acetate buffer adjusts to addition of a small amount of NaOH is:
- CH3COOH(aq) + H2O(1) → H30+ (aq) + CH3COO (aq)
<h3>What is a buffer?</h3>
A buffer is a solution which resists changes to its pH when small amounts of strong base or acid is added to it.
Buffers are made from solutions of weak acids and their salts or weak bases and their salts.
The equilibria showing how a buffer made from acetic acid and sodium acetate (NaCH3COO) adjusts to addition of a small amount of NaOH is as follows:
- CH3COOH(aq) + H2O(1) → H30+ (aq) + CH3COO (aq)
Addition of NaOH, a strong base will neutralize the hydronium ion, causing the acetic acid ionization equilibrium to shift to the right to produce more of the acetate ion, the conjugate base.
Learn more about acetate buffer at: brainly.com/question/17490438
Hello.
balance the equation first
<span>Cu + 2AgNO3 ----> [2Ag] + Cu(NO3)2
</span>
<span>Now set up the equation </span>
<span>350gCu*(1molCu/ 63.546g Cu) * (2molAg/1molCu) * (107.8682g Ag/1molAg)
</span>
<span>Cancel out 1molCu with 1molCu, and cancel out 2molAg with 1molAg (it still remains 2mol Ag) </span>
<span>You are left with </span>
<span>350gCu*(1/63.546g Cu)*(2Ag/1)* (107.8682g Ag/1)
</span>
<span>350gCu*1*2*107.8682= 75507.74g
</span>
<span>now divide 75507.74g by 63.546g
</span>
<span>you get [1188.2374972460896g Ag]
</span>
Have a nice day
