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Katen [24]
3 years ago
13

Compute 4.659×104−2.14×104. Round the answer appropriately

Chemistry
1 answer:
Vilka [71]3 years ago
7 0
 the answer is 261.976
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In the laboratory you dissolve 12.3 g of chromium(ii) sulfate in a volumetric flask and add water to a total volume of 375. ml.
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Convert, using dimensional analysis, from 24 days to seconds.
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How much heat is needed to raise a 0.30 g piece of aluminum from 30.C to 150C? The specific heat of aluminum 0.900j/gC
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8 0
3 years ago
Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)
Digiron [165]

Answer: =176.6kJmol^{-1}

Explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of  C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

\Delta H= {\text {sum of bond energies of reactants}}-  {\text {sum of bond energies of products}}

\Delta H= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

\Delta H= {1B.E(835kJmole^{-1})+2B.E(414kJmole^{-1}) +1B.E(436.4kJmole^{-1})} -  {1B.E(620kJmole^{-1})+4B.E(414kjmole^{-1})}

=176.6kJmol^{-1}







7 0
4 years ago
A 5.00 mL of a salt solution of unknown concentration was mixed with 35.0 mL of 0.523 M AgNO3. The mass of AgCl solid formed was
Lapatulllka [165]

Answer:

The chemical term in the equation for the precipitate of AgCl(s) is n=3.54*10^-3

Explanation:

the quantity of AgCl(s) in moles is:

n = 0.508g / 143.32 g/mol = 3.54*10^-3 mol

to verify it the mass of AgNO3 involved in the reaction should be

n AgNO3 required = n = 3.54*10^-3 mol

the mass of n involved should be higher than n AgNO3

n existing = V*N = 0.523 mol/L * 35*10^-3 L = 18.305*10^-3 mol

5 0
3 years ago
Read 2 more answers
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