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3 years ago

When two substances react to form products, the reactant which is used up is called the ____.

1 answer:

3 0

The reactant being used up is called limiting reagent as it limits the total amount of product produced.

if 4 units of HCL gives 2 units of Cl therefore
4:2
0.98:x
x=(0.98*2) /4
x=0.49L

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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:

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

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3 years ago

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Its a chemical reaction because both are affected by the reaction and changed. hope this helps (ू• o •ू )

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Answer:

0.025 g C6H12O6

Explanation:

ppm = (g solute/ g solution)* 10^6

g solute= (ppm * g solution)/ 10^6

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g solute=0.025 g C6H12O6

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liam uses a machine that counts waves on a metal cable. The machine counts 25 waves in 2 seconds. what is the frequency of the w

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Frequency is defined as the number of waves per second. In this machine 25 waves pass in one second.
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