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3 years ago

List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)

Chemistry

1 answer:

Leokris [45]3 years ago

3 0

Answer:

20, 44, 62

Explanation:

To find the number of atoms of each element, we multiply coefficient and subscript

For example $5 Ca_1 Cl_2$ contains

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example $3Ca_3 (P_1 O_4 )_2$

we multiply coefficient × subscript × number outside the bracket to find the number of atoms

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

3 × 3 = 9 , Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present.

So let us find the number of atoms of each element on the left side of the equation

$2C_{10} H_{22} (l)+31O_2 (g)\Rightarrow 20CO_2 (g)+22H_2 O(g)$

Number of C atoms = 2 × 10 = 20

Number of H atoms = 2 × 22 = 44

Number of O atoms = 31 × 2 = 62

20, 44, 62 are the Answers.

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Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

3 years ago

The gas in a closed container has a pressure of 3.00 x 10² kPa 30 ° C. What will the pressure be if the temperature is lowered t

Rainbow [258]

Answer: 100kPa

Explanation:

P1 = 3.00 x 10² kPa , P2 =?

T1 = 30°C = 30 +273 = 303k

T2 = —172°C = —172 + 273 = 101k

P1/T1 = P2/T2

3 x 10² / 303 = P2 / 101

P2 = (3 x 10² / 303) x 101

P2 = 100kPa

6 0

3 years ago

Which of the following represents the least number of molecules?

lozanna [386]

Answer:

A ; 20g of water has the least number of molecules

Explanation:

Here, we want to know which of the options represent the least number of molecules;

To calculate this, we are going to first calculate the number of moles using the formula below;

Number of moles = mass/molar mass

Then from the number of moles, we can calculate the number of molecules by multiplying the number of moles by 6.02 * 10^23 molecules

Let’s proceed;

a. That will be 20/18.02 = 1.1099 moles

1.1099 * 6.02 * 10^23 = 6.68 * 10^23 molecules

b. That will be 77/16.06 = 4.8 moles

4.8 * 6.02 * 10^23 = 2.89 * 10*24 molecules

c. That will be 68/42.09 = 1.62 moles

1.62 * 6.02 * 10^23 = 9.73 * 10^23 molecules

d. That will be 100/44.02 = 2.27 moles

2.27 * 6.02 * 10^23 = 1.37 * 10^24 molecules

e. That will be = 84/20.01 = 4.2 moles

4.2 * 6.02 * 10^23 = 2.53 * 10^24 molecules

From all the values obtained, the lowest is 20g of water

6 0

3 years ago

Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict wheth

Elena-2011 [213]

Answer:

$\Delta S_{surr} = + 0.32113\: kJ/K$

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at constant pressure is given by,

$\Delta S_{surr} = \frac{-\Delta H}{T}$

$\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K} \right ) = + 0.32113\: kJ/K > 0$

In the given reaction:

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

$\therefore \Delta S_{system} < 0$

As we know, for a spontaneous process, that the total entropy should be positive.

$\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0$

Therefore, at the given temperature,

if $\Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0$ then the given reaction is spontaneous

if $\Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0$ then the given reaction is non-spontaneous

6 0

3 years ago

Name the separation technique illustrated in the picture

o-na [289]

The picture isn’t included

3 0

2 years ago