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3 years ago

Which is larger aliter or a quart?

Chemistry

2 answers:

omeli [17]3 years ago

7 0

Answer:

a liter is larger, hope that helps :)

just olya [345]3 years ago

7 0

Answer:

1 us liquid quart =

0.946 litres

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Describe the heat involved if you touch a flask containing an on-going endothermic reaction

juin [17]

The flask would feel cooler than when the reaction first started.

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3 years ago

How many grams of oxygen are needed to produce 5.40 g of water according to this equation? 2 H2 (g) + O2 (g) ⟶ 2 H2O (l)

garik1379 [7]

Explanation:

https://socratic.org/questions/using-the-equation-2-h2-o2-2-h2o-determine-how-many-grams-of-oxygen-will-be-need

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2 years ago

A chemistry student needs of acetone for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student dis

galben [10]

Answer:

The answer is " $7.90 \ g$ ".

Explanation:

Please finds the complete question in the attached file.

$Volume = 10.0 \ mL \\\\Density =0.79 g\ cm^3$

$\text{Density}=\frac{Mass}{Volume}$

$Mass = Density \times Volume$

$=(0.790 g\ cm^3) \times 10.0 \ mL\\\\=7.90 \ g$

The mass of acetone to be weighed is $7.90 \ g$

8 0

2 years ago

How to Balance __MgF2 + __Li2CO3 + __ LiF

Andru [333]

__MgF2 + __Li2CO3 + __ 2LiF

7 0

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!

3 0

3 years ago