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strojnjashka [21]
3 years ago
12

What is the final chemical equation from the following intermediate chemical equations?

Chemistry
1 answer:
gavmur [86]3 years ago
5 0

Answer:

The final chemical equation would be :

A.

P_{4}O_{6}+2O_{2}\rightarrow P_{4}O_{10}

Explanation:

This is obtained by adding the respective intermediates:

P_{4}O_{6}\rightarrow P_{4} + 3O_{2}

P_{4}+5O_{2}\rightarrow P_{4}O_{10}

P4(s) are on opposite side of arrow so cancel each other.(No P4 in final reaction)

O2(g) = 5 - 3 = 2

There are 2 O2 more on left (reactant side)

finally there are ,

P4O6 and 2 O2 on left hand side and P4O10 on Right side of the arrow

So we get

P_{4}O_{6}+2O_{2}\rightarrow P_{4}O_{10}

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What do the superscripts and subscripts in the notation ^40 K represent?
raketka [301]

Explanation:

subscript is K

superscript is ^

subscript K means a unit of temperature like F or C

superscript ^ means to the power of

So all together it means to the power of 40 K

3 0
3 years ago
How much potassium chloride is needed to make 0.500 m solution with 1.50 L of water?
Andrew [12]

Answer:

55.9 g KCl.

Explanation:

Hello there!

In this case, according to the definition of molality for the 0.500-molar solution, we need to divide the moles of solute (potassium chloride) over the kilograms of solvent as shown below:

m=\frac{mol}{kilograms}

Thus, solving for the moles of solute, we obtain:

mol=m*kilograms

Since the density of water is 1 kg/L, we obtain the following moles:

mol=0.500mol/kg*1.50kg\\\\mol=0.75mol

Next, since the molar mass of KCl is 74.5513 g/mol, the mass would be:

0.75mol*\frac{74.5513g}{1mol}\\\\55.9g \ KCl

Regards!

4 0
3 years ago
The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5
Natasha2012 [34]
For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
4 0
3 years ago
Controlled nuclear chain reactions
Lunna [17]
The nuclear reactions which are under experimenter's control are said to be controlled nuclear reactions. In this, you can maintain the speed of the incident particle. α and β-decay process are examples of non-controlled nuclear reactions. 
8 0
3 years ago
Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, w
puteri [66]

Answer:

= -356KJ

<em>therefore, the reaction where heat is released is exothermic reaction since theΔH is negative</em>

Explanation:

given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ

then,

change in enthalpy (ΔH) = -162( released energy)

work(w) = -194KJ

change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.

At constant pressure the change in enthalpy is given as

ΔH = ΔU + PΔV

ΔU = change in energy

ΔV = change in volume

P = pressure

w =  -pΔV

therefore,

ΔH = ΔU -W

to evaluate  energy change we have,

ΔU =ΔH + W

ΔU = -162+ (-194KJ)

= -356KJ

<em>therefore, the reaction where heat is released is exothermic reaction since theΔH is negative</em>

6 0
3 years ago
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