What about matter can change and what does not change according to the law of conservation?

If 8.700 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 Â°c, what is the final tem

leonid [27]

C₆H₆ is benzene which has a molar mass of 78 g/mol. When benzene is burned, the reaction is called combustion. The heat produced in this reaction is called the heat of combustion. For benzene, the heat of combustion is -3271 kJ/mol.

Heat of benzene = (8.7 g)(1 mol/78 g)(-3271 kJ/mol) = -364.84 kJ

By conservation of energy,
Heat of benzene = - Heat of water
where
Heat of Water = mCp(Tf - T₀)
where Cp for water is 4.187 kJ/kg·°C

Thus,

-364.84 kJ = -(5691 g)(1 kg/1000 g)(4.187 kJ/kg·°C)(Tf - 21)
Tf = 36.31°C

8 0

3 years ago

1. Burning coal and oil in a power plant produces pollutants, such as sulfur dioxide, SO2. This compound can be removed by the f

Lorico [155]

Idk tbh srry man am in quarantine so I have a big packet to do

3 0

3 years ago

Write balance complete molecular equation, ionic equation, and net ionic equations for the reactions that occur when each of the

AveGali [126]

Answer: The complete molecular, ionic, and net ionic equations are given below. The spectator ions are sodium and nitrate ions.

Explanation:

The ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

The net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

The balanced molecular equation for the reaction of lead (II) nitrate and sodium sulfide follows:

$Pb(NO_3)_2(aq)+Na_2S(aq)\rightarrow PbS(s)+2NaNO_3(s)$

The ionic equation follows:

$Pb^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow PbS(s)+2Na^+(aq)+2NO_3^-(aq)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+S^{2-}(aq)\rightarrow PbS(s)$

3 0

3 years ago

How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh? h3po4 + koh yields h2o + k3po4 be sur

schepotkina [342]

the balanced chemical equation for the reaction is as follows

H₃PO₄ + 3KOH ---> K₃PO₄ + 3H₂O

stoichiometry of H₃PO₄ to KOH is 1:3

first we have to find which the limiiting reactant is

as the amount of product formed depends on the amount of limiting reactant present

number of H₃PO₄ moles reacted - 4.4 mol

if H₃PO₄ is the limiting reactant

1 mol of H₃PO₄ reacts with 3 mol of KOH

then 4.4 mol of H₃PO₄ reacts with - 3 x 4.4 mol = 13.2 mol of KOH

but only 3.8 mol of KOH is present

therefore KOH is the limiting reactant

stoichiometry of KOH to K₃PO₄ is 3:1

number of KOH moles reacted - 3.8 mol

therefore number of K₃PO₄ formed = number of KOH moles reacted / 3

= 3.8 mol / 3 = 1.3 mol

answer is 1.3 mol of K₃PO₄

3 0

3 years ago

If you burn 48.6 g of hydrogen and produce 434 g of water, how much oxygen reacted?

GarryVolchara [31]

Answer:

385.69 g of O₂

Solution:

The Balance Chemical equation for said reaction is as follow;

2 H₂ + O₂ → 2 H₂O

According to Equation,

4.032 g ( 2 mol) H₂ reacts to produce = 36.03 g (2 mol) of H₂O

So,

48.6 g H₂ on reaction will produce = X g of H₂O

Solving for X,

X = (48.6 g × 36.03 g) ÷ 4.032 g

X = 434.29 g of H₂O

It means that the H₂ provided is in Excess. Therefore, the yield of product (H₂O) is being controlled by O₂ (Limiting Reagent).

So, According to Equation,

36.03 g (2 mol) H₂O is produced by = 31.998 g (1 mol) of O₂

So,

434.29 g of H₂O will be produced by = X g of O₂

Solving for X,

X = (434.29 g × 31.998 g) ÷ 36.03 g

X = 385.69 g of O₂

8 0

3 years ago