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Soloha48 [4]
2 years ago
15

Balance these equations! 1) AIBr; + K- KBr + AI

Chemistry
1 answer:
Tema [17]2 years ago
3 0

Answer:

Instructions on balancing chemical equations:

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below

Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.     Compare: Co - cobalt and CO - carbon monoxide

To enter an electron into a chemical equation use {-} or e

To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.

Example: Fe{3+} + I{-} = Fe{2+} + I2

Substitute immutable groups in chemical compounds to avoid ambiguity.

For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,

but PhC2H5 + O2 = PhOH + CO2 + H2O will

Compound states [like (s) (aq) or (g)] are not required.

If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.

Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reage

Explanation:

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Answer:

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6 0
2 years ago
the equation demonstrates a reaction of sodium chloride and fluorine. 2NaCl+F2=2NaF+Cl2. in order for the reaction to proceed, f
slamgirl [31]

Answer: You would need 1 mole of Fluorine

Explanation:The equation is already balanced so just looking at the coefficients in the equation we can see that Sodium Chloride (2NaCl) needs two moles for this equation and fluorine (F2) only needs one.

6 0
2 years ago
Which of the following is not a trait of gases?
iVinArrow [24]
<span>d. tightly packed particles</span>
7 0
3 years ago
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The Connecticut Bueller theory assumes that particles of an ideal gas are
postnew [5]

<span>are in random, constant, straight-line motion</span>

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3 0
2 years ago
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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