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2 years ago

13

How many grams of Cu will be produced if 15. 8 grams of Cu2S got consumed?

1 answer:

White raven [17]2 years ago

7 0

Answer:

15

Explanation:

Because the ratio of copper in the solution is 15

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How many kilograms of NH3 are needed to produce 1.90x10^5 of (NH4)2SO4? Answer in scientific notation

leva [86]

Molar mass :

NH₃ = 17.0 g/mol

(NH₄)₂SO₄ = 132 g/mol

2 NH₃ + H₂SO₄ = (NH₄)₂SO₄

2 x 17 g NH₃ ------------ 132 g <span> (NH</span>₄)₂SO<span>₄
</span>( mass NH₃) ------------ 1.90 x 10⁵ <span> (NH</span>₄)₂SO<span>₄

( mass NH</span>₃) = (1.90 x 10⁵) x 2 x 17 / 132
<span>
( mass NH</span>₃) = 6460000 / 132
<span>
( mass NH</span>₃) = 48939.39 g
<span>
in kg :

48939.39 / 1000 = 48.93 kg = 4.893 x 10</span>¹ kg
<span>
hope this helps!
</span><span>
</span>

6 0

3 years ago

Give the value of the quantum number ℓ, if one exists, for a hydrogen atom whose orbital angular momentum has a magnitude of 6√(

MakcuM [25]

Answer:

For this angular momentum, no quantum number exist

Explanation:

From the question we are told that

The magnitude of the angular momentum is $L = 6\sqrt{\frac{h}{2 \pi} }$

The generally formula for Orbital angular momentum is mathematically represented as

$L = \sqrt{(l * (l + 1)) } * \frac{h}{2 \pi}$

Where $l$ is the quantum number

now

We can look at the given angular momentum in this form as

$L = 6\sqrt{\frac{h}{2 \pi} } = \sqrt{36} * \sqrt{\frac{h}{2 \pi}} }$

comparing this equation to the generally equation for Orbital angular momentum

We see that there is no quantum number that would satisfy this equation

5 0

3 years ago

kozerog [31]

Answer:

Average of measurements = 20.97 cm

Explanation:

Given data:

Three measurements of pete's = 20.9 cm, 21.0 cm, 21.0 cm.

Average value of measurements = ?

Formula:

Average of measurements = sum of all measurements / Total number of measurements

Solution:

Average of measurements = 20.9 cm + 21.0 cm + 21.0 cm. / 3

Average of measurements = 62.9 cm / 3

Average of measurements = 20.97 cm

7 0

3 years ago

What is the highest energy to which doubly ionized helium atoms (alpha particles) can be accelerated in a DC accelerator with 3

pentagon [3]

Answer:

Highest energy will be equal to $9.6\times 10^{-13}J$

Explanation:

Charged on doubly ionized helium atom $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

It is accelerated with maximum voltage of 3 MV

So voltage $V=3\times 10^6volt$

Now energy is given by $E=qV=3.2\times 10^{-19}\times 3\times 10^6=9.6\times 10^{-13}J$

So highest energy will be equal to $9.6\times 10^{-13}J$

5 0

3 years ago

The chemical equation of photosynthesis includes 6O2. What does the small number after the O represent?

blondinia [14]

Its the number of Oxygen molecules

3 0

3 years ago