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2 years ago

¿Cuántos moles hay en 350 g de NaNO3?

Chemistry

1 answer:

olga55 [171]2 years ago

4 0

Answer:

3 moles

Explanation:

So, approximately 3 moles of NaNO3 can be obtained, by reacting 253 grams of Na2CrO4. Also, the number of moles is a dimensionless quantity.

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What period is the atom in?

lys-0071 [83]

It is in period 3
It is in group 17
It is a chlorine atom because it has 17 electrons which means the atomic number is 17

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3 years ago

Spontaneous generation is a mistaken idea because living things:

Oduvanchick [21]

B. Are produced only by living things

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3 years ago

Read 2 more answers

How many grams of iron oxide, Fe2O3 will be produced if 165 g of O2 gas is supplied? (follow the same steps as mol to mol, only

forsale [732]

Answer:

$m_{Fe_2O_3}=549gFe_2O_3$

Explanation:

Hello there!

In this case, according to the given chemical reaction for this problem about stoichiometry:

$4Fe+3O_2\rightarrow 2Fe_2O_3$

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

$m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3} \\\\m_{Fe_2O_3}=549gFe_2O_3$

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3 years ago

What compound do cells break down for energy in the process of cellular respiration?

arsen [322]

Answer:

Glucose

Explanation:

Glucose is broken down into water and carbon dioxide.

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6 0

3 years ago

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate

bekas [8.4K]

Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)$

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

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3 years ago