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2 years ago

Which of the following energy quantities is equivalent to 578 J?

Chemistry

1 answer:

svlad2 [7]2 years ago

5 0

Answer:

0.138 kcal

Explanation:

The answer choices are:

a) 5.78 × 10⁵ kJ

b) 138 kcal

c)0.138 kcal

d)1.38 × 10⁵ kcal

<h2>Solution</h2>

You must use the conversion factors to convert J to kcal.

1cal = 4.184J

Dividing both sides by 4.184J:

⇒ 1 = 1cal/4.184J

1kcal = 1,000cal

Dividing both sides by 1,000 cal

⇒ 1 = 1kcal / 1,000 cal

Mutiply 578J by the conversion factors to find kcal:

$578J\times 1cal/4.184J\times 1kcal/1,000cal=0.138kcal$

Notice how the units J and cal cancel to yield kcal at the end.

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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c

Pepsi [2]

Answer:

The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)

Explanation:

We make use of the formula

$Q=m \times c \times \Delta T$

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

$Q=m \times c \times \Delta T$

Rearranging the formula

$c= \frac {Q}{(m\times \Delta T)}$

$= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}$

So,

c = 0.129 J/(g℃)

(Answer)

7 0

2 years ago

500ml of a buffer solution contains 0.050 mol nahso3 and 0.031

nydimaria [60]

Answer:

The answers are explained below

Explanation:

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.

4 0

2 years ago

According to the following reaction, what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of 0.210 M Pb

svetoff [14.1K]

Answer:

86 mL

Explanation:

First find the moles of Pb (NO3)2

n=cv

where

c ( concentration)= 0.210 M

v ( volume in L) =0.05

n= 0.210 × 0.05

n= 0.0105

Using the mole ratio, we can find the moles of KCl by multiplying by 2

n (KCl) =0.0105 ×2

=0.021

v (KCl)= n/ c

= 0.021/ 0.244

=0.08606557377

=0.086 L

= 86 mL

8 0

2 years ago

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Two protons and two neutrons are released as a result of this reaction.

Murljashka [212]

Answer: The particle released in the given reaction is one alpha particle

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{222}_{86}\textrm{Rn}\rightarrow ^A_Z\textrm{X}+^{218}_{84}\textrm{Po}$

To calculate A:

Total mass on reactant side = total mass on product side

222 = A + 218

A = 4

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

86 = Z + 84

Z = 2

The isotopic symbol of unknown element is $_{2}^{4}\textrm{He}$ . Another name of helium atom is alpha particle.

Hence, the particle released in the given reaction is one alpha particle

3 0

2 years ago

Read 2 more answers