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xenn [34]
3 years ago
10

A reaction produces 0.863 moles of H2O. How many molecules are produced?

Chemistry
2 answers:
Stolb23 [73]3 years ago
8 0
Number of molecules = moles x 6.02 x 10^23 = 0.863 x  6.02 x 10^23 
                                                                         =    5.19 x 10^23 molecules.
balu736 [363]3 years ago
4 0
To convert from moles to molecules, you multiply by Avogadro's constant.

0.863 moles H₂O * (6.022*10^23 molecules/1 mole) = 5.20*10^23 molecules H₂O

5.20*10^23 molecules of H₂O are produced.
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its either A or D

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3 years ago
Our main source of energy is the sun. <br><br><br><br> True or False
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Our main source of energy is the sun. True.

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3 years ago
What is the nuclear binding energy of an atom that has a mass defect of 5.0446
notsponge [240]

Answer:

<em>Option C: 4.54 x </em>10^{11}<em> KJ/mol of nuclei</em>

<em>Note: </em>Here in this question option C is not correctly put. It is 4.54 x10^{11} rather than 4.54 x 10^{-123}.

Explanation:

If mass defect is known, then nuclear binding energy can easily be calculated, here's how:

First step is to convert that mass defect into kg.

Mass defect = 5.0446 amu

Mass defect = 5.0466 x 1.6606 x 10^{-27}

Because 1 amu = 1.6606 x 10^{-27} Kg.

<em>Mass defect = 8.383 x </em>10^{-27}<em> kg.</em>

Now, we need to find out it's energy equivalent by using following equation:

Using the equation E = mc²:

where c= 3.00 x 10^{8} m/s²

E = (8.383 x 10^{-27}) x (3.00 x 10^{8})²

E = 7.54 x 10^{-10} J  this energy is in Joules but nuclear binding energy is usually expressed in KJ/mol of nuclei. Let's convert it:

(7.54 x 10^{-10} Joule/nucleus)x(1 kJ/1000 Joule)x(6.022 x 10^{23} nuclei/mol) =  

<em>4.54 x </em>10^{11}<em> kJ/mol of nuclei .</em>

E = <em>4.54 x </em>10^{11}<em> kJ/mol of nuclei .</em> So, this is the nuclear binding energy of that atom, which is option  C.

<em>Note:</em> Here in this question option C is not correctly put. It is 4.54 x10^{11} rather than 4.54 x 10^{-123}

4 0
3 years ago
Read 2 more answers
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
blsea [12.9K]
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

                    =0.00516 M 

Thus, pOH = 2.30 
 pH = 14 - pOH = 11.7
6 0
3 years ago
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