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sesenic [268]
3 years ago
14

Nomenclature for NaC2H3O2

Chemistry
1 answer:
adelina 88 [10]3 years ago
7 0
That would be Sodium acetate
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ASAP I AM GIVING BRAINLIEST PLEASEEEEEEEEE HELPPPPPPPPPPPPPPPPPP
zepelin [54]

Answer:

B?

Explanation:

In the example, the amount of hydrogen is 202,650 x 0.025 / 293.15 x 8.314472 = 2.078 moles. Use the mass of the hydrogen gas to calculate the gas moles directly; divide the hydrogen weight by its molar mass of 2 g/mole. For example, 250 grams (g) of the hydrogen gas corresponds to 250 g / 2 g/mole = 125 moles.

7 0
3 years ago
How many moles are there in 24.00 g of NaCl
Elis [28]

Answer:

The answer to your question is 0.41 moles

Explanation:

Data

moles of NaCl = ?

mass of NaCl = 24 g

Process

To solve this problem just calculate the molar mass of NaCl, and remember that the molar mass of any substance equals to 1 mol.

1.- Calculate the molar mass

NaCl = 23 + 35.5 = 58.5 g

2.- Use proportions and cross multiplication

               58.5 g of NaCl ------------------- 1 mol

               24.0 g               ------------------- x

                     x = (24 x 1) / 58.5

                     x = 0.41 moles

6 0
3 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
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Which ones are false? need help please.
kirill [66]

Answer:

A.) False B.)False C.) True D.) False E.) False

Explanation:

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3 years ago
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It would take one light year to travel to the sun true or false?​
slava [35]
True! I’m pretty sure
3 0
2 years ago
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