Which chemical symbols will complete the equation for this single displacement reaction?

How do you know that a sealed calorimeter is a closed system?

3241004551 [841]

Answer: Option (c) is the correct answer.

Explanation:

When a system is open then there will be exchange of energy between the system and surrounding.

Whereas when a system is closed then there will be no exchange of energy, that is, thermal energy will not flow into the atmosphere.

Thus, we can conclude that a sealed calorimeter is a closed system because thermal energy is not transferred to the environment.

5 0

3 years ago

Read 2 more answers

A racing car travels 340 miles in 120 minutes. what speed is traveling at?

Papessa [141]

Answer:

It is traveling at 170 miles per hour.

Explanation:

divide 340 by 2 because right now it's at 340 miles per two hours.

You get 170 miles per hour.

4 0

2 years ago

890J of heat are applied to a piece of aluminum, causing a 4.6°C increase in its temperature. The specific heat of aluminum is 0

Semmy [17]

Answer:

The answer would be C 214g

Explanation:

890j of heat causes 4.6°c increase in temperature

specific heat of aluminium after is o.9022 j /g°c

now by using the formula .The mass of aluminium would be c that is 214 g

7 0

3 years ago

Metric dimensional analysis worksheet problem: 645 kg to g

Andrews [41]

Answer:

645000 grams my dude

3 0

3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

3 years ago