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PSYCHO15rus [73]
3 years ago
15

Gold and hydrochloric acid

Chemistry
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

hydrochloric acid can be used to clean gold

hydrochloric acid does not gold in any way

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Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com
lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30  = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0
3 years ago
Which the following statement ia correct for the equation shown here​
Shkiper50 [21]
B.

As you can see both NO and NH3 have 4 moles therefore it is 4:4 between the molecules or in other words a 1:1 ratio in simplest forms
7 0
3 years ago
A machine
rosijanka [135]

i believe the answer is A (((:

4 0
2 years ago
Read 2 more answers
Which is the smallest part of a compound called?
lidiya [134]
The answer is molecule

6 0
3 years ago
Read 2 more answers
1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?
love history [14]

Answer:

m=1.325gNa_2CO_3

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

eq-g=0.5eq-g/L*0.050L=0.025eq-g

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3

Finally, the mass is computed by using its molar mass (106 g/mol)

m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3

Regards.

7 0
3 years ago
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