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3 years ago

10) How many grams are there in 1.00 x 10^24 molecules of BC13?

Chemistry

1 answer:

lana66690 [7]3 years ago

3 0

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

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3 years ago

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A 13-gram rubber stopper is attached to a 0.93-meter

makkiz [27]

<h3>Answer:</h3>

The centripetal acceleration is 26.38 m/s²

<h3>Explanation:</h3>

We are given;

Mass of rubber stopper = 13 g
Length of the string(radius) = 0.93 m
Time for one revolution = 1.18 seconds

We are required to calculate the centripetal acceleration.

To get the centripetal acceleration is given by the formula;

Centripetal acc = V²/r

Where, V is the velocity and r is the radius.

Since time for 1 revolution is 1.18 seconds,

Then, V = 2πr/t, taking π to be 3.142 ( 1 revolution = 2πr)

Therefore;

Velocity = (2 × 3.142 × 0.93 m) ÷ 1.18 sec

= 4.953 m/s

Thus;

Centripetal acceleration = (4.953 m/s)² ÷ 0.93 m

= 26.38 m/s²

Hence, the centripetal acceleration is 26.38 m/s²

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3 years ago

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha

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There are 2.32 x 10^6 kg sulfuric acid in the rainfall.

Solution:
We can find the volume of the solution by the product of 1.00 in and 1800 miles2:
1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m
1.00 in * 1 m / 39.3701 in = 0.0254 m
Volume = 4.662 x 10^9 m^2 * 0.0254 m
= 1.184 x 10^8 m^3 * 1000 L / 1 m3
= 1.184 x 10^11 Liters

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70:
[H+] = 10^-pH = 10^-3.7 = 0.000200 M
[H2SO4] = 0.000100 M

By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid:
1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4

We can now calculate for the mass of sulfuric acid in the rainfall:
mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
= 2.32 x 10^9 g * 1 kg / 1000 g
= 2.32 x 10^6 kg H2SO4

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4 years ago

Read 2 more answers

Car manufacturers are developing engines that use H₂ as fuel. In Iceland, Sweden, and other parts of Scandinavia, where hydroele

weeeeeb [17]

If the coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.

(a) The reaction is

2H₂O + 2e⁻ → H₂ + 2OH⁻

According to the reaction, 2 moles of electrons flow per mole of reaction,

(b) Given

T = 25°C

Change Celsius into Kelvin

T = 25°C

= (25 + 273 ) K

= 298 K

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It is expressed as

PV = nRT

Hence ,

n = $\frac{PV}{RT}$

= $\frac{(12)(3.5.10^{6} )}{(0.082057)(298)}$

= 1.7176 ×10⁶ mole

Now, From the given reaction we can say that ,1 mole of H₂ is produced by 2 moles of electron

Hence, 1.7176 × 10⁶ mole of H₂ is produced by

= 2 × 1.7176 × 10⁶ moles of electron

= 3.435 ×10⁶ moles of electron

We know that

charge = moles of electron × F

= (3.435 ×10⁶ × 96500) C

= 3.31 × 10¹¹ C

Voltage = $\frac{energy}{charge}$ = $\frac{J}{C}$

1.44 V = $\frac{energy}{3.31 . 10^{11} }$

Energy = 1.44 V × 3.31 × 10¹¹ C

= 4.766 × 10¹¹ Joules

Thus from the above conclusion we can say that, coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.

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c) If the coulombs are supplied at 1.44 V, how many joules are produced?

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2 years ago

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Answer:

Since both components of these solutions have the same molar mass, mole fractions would be the same as mass fractions.

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4 years ago