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Mashcka [7]
3 years ago
9

if you supply 3600 kJ of heat, how many grams of ice at 0°C can be melted, heated to its boiling point? (Make M your X in equati

ons) (figure out which q you need)
Chemistry
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

=1.36kg

Explanation:

First heat is absorbed to melt the ice- latent heat of fusion of ice then heat is absorbed to raise the temperature to 373K

The specific heat capacity of water is 4.187kJ/kg/K while  the latent heat of fusion of ice is 2230kJ/kg/K.

Letting the mass of the ice to be =x

x×2230kJ/kg/K + x×4.187kJ/kg/K×100K= 3600kJ

2230x+418.7x=3600

2647.8x=3600

x=1.36 kg

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What is the relationship between the strength of the magnetic field and the deflection of the galvanometer needle?
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1 year ago
Which statement is true about the exothermic reaction?
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C

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3 years ago
Cuando se trata Ca3P2 con agua, los productos son Ca(OH)2 y PH3. Calcular el peso máximo obtenido al reaccionar 2 g de Ca3P2 con
anyanavicka [17]

Answer:

0.629 gramos de PH3 es la máxima de cantidad que puede ser producida.

Explanation:

¡Hola!

En este caso, dado que tenemos la siguiente reacción química, la cual se puede balancear directamente:

Ca_3P_2+6H_2O\rightarrow 3Ca(OH)_2+2PH_3

Podemos calcular la masa máxima de cualquier producto, digamos PH3, al comparar la masa de este, que 2 g the Ca3P2 y 1 g de H2O producen por separado y de acuerdo a la estequiometría:

2gCa_3P_2 *\frac{1molCa_3P_2 }{182.18gCa_3P_2 }*\frac{2molPH_3}{1molCa_3P_2 }  *\frac{34gPH_3}{1molPH_3 } =0.747gPH_3\\\\1gH_2O *\frac{1molH_2O }{18.02gH_2O}*\frac{2molPH_3}{6molH_2O}  *\frac{34gPH_3}{1molPH_3 } =0.629gPH_3

De este modo, infermos que solamente 0.629 gramos the PH3 pueden ser obtenidos al ser el agua el reactivo límite.

¡Saludos!

5 0
3 years ago
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