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3 years ago

Help please the questions are confusing

Chemistry

2 answers:

Virty [35]3 years ago

7 0

It has 11 Electrons because Sodium is the normal atom, Electrons are negative, Protons are positive. If there are 11 protons there must be 11 electrons to cancel each other out

SashulF [63]3 years ago

3 0

Answer:

5- number of electrons=11

Explanation:

in a neutral atom number of protons=number of electrons which in this case=11

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A sky rocket is launched with a force of 10 N and accelerates at 20 m/s squared what is the mass of this skyrocket

tia_tia [17]

Answer: 0.5 kg

Explanation:

Force of launched skyrocket = 10 N

Acceleration of skyrocket = 20 m/s squared (i.e 20m/s^2)

Mass of the skyrocket = ?

Recall that Force is the product of the mass of an object by the acceleration by which it moves.

i.e Force = Mass x Acceleration

10N = Mass x 20m/s^2

Mass = (10N/20m/s^2)

Mass = 0.5 kg

Thus, the mass of this skyrocket is 0.5 kilograms

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3 years ago

sweet-ann [11.9K]

Answer:lmk when u have the answer

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2 years ago

Calculate the percentage of all elements in Fel(OH)3

Vsevolod [243]

Answer: 30.06% is your answer.

I hope this helps.

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2 years ago

The total mass of reactants and products is _______ during a chemical reaction

SIZIF [17.4K]

Answer: The total mass of reactants and products is the same during a chemical reaction.

Explanation:

the total mass of reactants is equal to the total mass of products.

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2 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago