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4 years ago

Use the periodic table to complete each nuclear fusion equation.

Chemistry

2 answers:

Verdich [7]4 years ago

6 0

Answer: A:5 B:2 C:15 D:8 E:O

Explanation:

just checked on edg

Misha Larkins [42]4 years ago

5 0

Answer:

a:5, b:2, c:15, d:8, e:O

For e its the letter o not zero

Explanation:

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Collisions may occur between tectonic plates. The resulting geological features differ depending on whether the colliding plates

nika2105 [10]

B. Amid - ocean ridge is formed

8 0

3 years ago

Calculate the number of moles of C in the hydrocarbon.

pochemuha

Answer:

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Explanation:

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6 0

3 years ago

What is the molarity of NaCL solution that has 58 grams of NaCL dissolved in 4.0 liters of water

Anna007 [38]

14.5 M

Molarity=grams/liters

M=58/4.0

4 0

3 years ago

Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behav

kvv77 [185]

Answer :

'n' specifies → (B) The energy and average distance from the nucleus.

'l' specifies → (C) The subshell orbital shape.

'ml' specifies → (A) The orbital orientation.

Explanation :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as $m_l$ . The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of

Spin Quantum number : It describes the direction of electron spin. This is represented as $m_s$ . The value of this is $+\frac{1}{2}$ for upward spin and $-\frac{1}{2}$ for downward spin.

As per question we conclude that,

'n' specifies → The energy and average distance from the nucleus.

'l' specifies → The subshell orbital shape.

'ml' specifies → The orbital orientation.

5 0

3 years ago

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0

3 years ago