<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it. Crude appearance. 18 18 0 F --> O + e 9 8 1 Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts. The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons. Followed by a right arrow to indicate the direction the reaction is going. Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons. Followed by a plus sign to indicate more. Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>

5 0

2 years ago

Read 2 more answers

A chemist uses hot hydrogen gas to convert chromium(III) oxide to

Darya [45]

Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$

Explanation:

The reaction equation for given reaction is as follows.

$Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O$

Here, 1 mole of $Cr_{2}O_{3}$ reacts with 3 moles of $H_{2}$ .

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide $(Cr_{2}O_{3})$ is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of $Cr_{2}O_{3}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol$

Now, moles of $H_{2}$ .given by 0.5 mol of $Cr_{2}O_{3}$ is calculated as follows.

$0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}$

As molar mass of $H_{2}$ is 2.016 g/mol. Therefore, mass of $H_{2}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g$

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$ .

7 0

3 years ago

How many atoms of Pb are there?

jeka94

Answer: i believe it is

602,000,000,000,000,000,000,000 atoms

Explanation:

6.02×1023 atoms/mole

6 0

2 years ago

How many moless is 4.91x10^22 molecules of H3PO4?

Lynna [10]

1 mol = 6.023x10^23 number of molecules (Avogadro's number)

1 : 6.023x10^23
X : 4.91x10^22

(6.023x10^23)X = 4.91x10^22

X = 4.91x10^22/6.023x10^23

X = 0.082 Moles

6 0

3 years ago