Ask question

Ask question

All categories

3 years ago

13

Element fact sheet Help!!!!

1 answer:

Alex17521 [72]3 years ago

7 0

Element name: Neon
element symbol: Ne
atomic number: 10
atomic mass: 20.1797
number of protons: 10
number of neutrons: 10
number of electrons: 2,8
interesting fact: the name means new, and it is pulled out of the air.
for the rest just add a neon bohr model ig

You might be interested in

What is the mass in milligrams of 4.30 moles of sodium? use significant figures?

alexandr1967 [171]

Answer is: mass <span>of 4,30 moles of sodium</span> is 98800 mg.
n(Na) = 4,30 mol.
m(Na) = ?
m(Na) = n(Na) · M(Na).
m(Na) = 4,30 mol · 23 g/mol.
m(Na) = 98,90 g.
m(Na) = 98,90 g · 1000 mg/1g.
m(Na) = 98900 mg.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.

5 0

3 years ago

The message that a neuron carries

Vinvika [58]

<span>The message that a neuron carries is the nerve. This gives a signal and acts as a nerve. </span>

6 0

3 years ago

A neutral atom of beryllium (Be) has an average mass of 9 amu and 4 electrons. How many neutrons does it have?

kodGreya [7K]

The answer is 5 subtract it

8 0

3 years ago

Read 2 more answers

What is the definition of the molar mass of an element or compound?

VashaNatasha [74]

Answer:

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance(mol) .

Explanation:

hope it may help you

mark as brainlist please

8 0

2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

Read 2 more answers