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zhuklara [117]
3 years ago
6

2Al + 6HCl → 2AlCl3 + 3H2 If 3.0 grams of H2 were produced, how many grams of Al reacted? (Molar mass of Al = 27 g/mol, molar ma

ss of H2 = 2.0 g/mol) ___ grams

Chemistry
1 answer:
Monica [59]3 years ago
8 0
Hope it cleared your doubt.

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What’s the answer pls hrlp
arsen [322]

Answer:

The answer is A

Explanation:

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3 years ago
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

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3 years ago
Some plz help me make mnemonic to helpme remember the oder of the planets​
Iteru [2.4K]

Answer:

My - Mercury

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Us- Uranus

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Explanation:

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3 years ago
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PLEASE HELP!!!
Nataly [62]
1. research question
2. background research
3. hypothesis
4. <span>Controlled experiment 
5. data analysis 
6. data collection
7. conclusion</span>
7 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!
pogonyaev

Answer:

m_{B}^{theoretical}=0.365gB

Y=87.1\%

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\   m_{B}^{theoretical}=0.365gB

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%

Best regards!

4 0
2 years ago
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