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3 years ago

How many grams of magnesium oxide forms when 8.0g of magnesium are burned?

1 answer:

7 0

Magnesium burns with oxygen to give magnesium oxide based on the following reaction:
2Mg+O2 ........> 2MgO(s)+energy
From the periodic table:
molecular mass of magnesium = 24.3 grams
molecular mass of oxygen = 16 grams

From the reaction:
2 x 24.3 = 48.6 grams of magnesium are required to produce 2(24.3+16) = 80.6 grams of magnesium oxide.
We can know the amount of magnesium oxide produced from 8 grams of magnesium by simply doing cross multiplication as follows:
amount of magnesium oxide = (8 x 80.6) / 48.6 = 13.267 grams

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During action potential, positively charged ________ ions move inside the cell.

Tems11 [23]

During action potential, positively charged sodium ions move inside the cell.

So option D is correct one.

The sodium ion moves inside the cell during a action potential. The stage of action potential is called depolarization . This open voltage gated sodium channel.

Action potentials ( those electrical impulse that send signals around body ) is nothing but more than temporary shift ( from negative to positive ) in the neuron's membrane potential caused by ions suddenly flowing in and out of the neuron.

It consists of phases:

Depolarization
overshoot
repolarization

An active potential propagates along the cell membrane of an axon until it reaches the terminal button.

to known more about action potential

brainly.com/question/27095019

#SPJ4

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2 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

What did GEM Anscombe stand for give a brief description?

KATRIN_1 [288]

Answer:

Explanation:

Anscombe is a philosopher who has approached contemporary themes, today enormously current, precisely in the ethical and social field, stimulating the development of virtue ethics as an alternative to utilitarianism, Kantian ethics and social contract theories. Anscombe argues that what we understand as a moral obligation, what we "must" do, is an obligation to God.

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3 years ago

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Sonja [21]

Answer:

ok 3 = 9 15 i got 15 so what

Explanation:

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3 years ago

PLEASE HELP ILL GIVE BRAINLIEST

tino4ka555 [31]

Answer:

False

Explanation:

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3 years ago

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