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-BARSIC- [3]
3 years ago
6

The equation NaCI + AgF -> NaF + AgCI is an example of what type of reaction

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Double replacement. The metal ions have exchanged partners.

“Single replacement” is <em>incorrect</em>. In single displacement, <em>one element displaces another element</em> in a compound.

“Decomposition” is <em>incorrect</em>. In decomposition, a <em>single substance</em> changes into two or more substances.

“Combustion” is <em>incorrect </em>because it usually involves a <em>reaction with oxygen</em> .

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stiks02 [169]
It comes from the core
8 0
3 years ago
Read 2 more answers
With white light, a team measured a 0.7% percent change with 3.5g of plant matter in a one liter container. Convert
Strike441 [17]

moles CO₂ = 5.57.10⁻⁴

<h3>Further explanation   </h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

Can be formulated :

\tt mol=\dfrac{mass}{MW}

0.7% percent change with 3.5g of plant matter

mass :

\tt 0.7\%\times 3.5~g=0.0245~g

moles :

\tt moles=\dfrac{0.0245}{44}=0.000557=5.57.10^{-4}

5 0
3 years ago
If you repeat an experiment and the result are very different from the result you got the first time the next step would be to w
Bess [88]
You would want to make sure that you have controlled the variables properly, and if you determine that you did then you would repeat the experiment to be sure of the results.
4 0
3 years ago
In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.68 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what
melomori [17]

Answer:

Y=65.7\%

Explanation:

Hello,

In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:

n_{MgI_2}^{theoretical}=2.68molMg*\frac{1molMgI_2}{1molMg} =2.68molMgI_2

Thus, we compute the percent yield as:

Y=\frac{n_{MgI_2}^{real}}{n_{MgI_2}^{theoretical}} *100\%=\frac{1.76mol}{2.68mol} *100\%\\\\Y=65.7\%

Best regards.

8 0
3 years ago
'Suppose you take 0.0332 kg of ammonium carbonate and dissolve it using 0.0395 kg of water. What would be the mass percent conce
VMariaS [17]

The solution is 45.7 % (NH₄)₂CO₃ by mass.

Mass of solution = 0.0332 kg + 0.0395 kg = 0.0727 kg

% (NH₄)₂CO₃ = Mass of (NH₄)₂CO₃/Total mass × 100 % = 0.0332 kg/0.0727 kg × 100 % = 45.7 %


8 0
3 years ago
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