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2 years ago

7

HELP FAST PLEASE! What is the limiting reactant if 43.4 g of NH3 react with 30 g of NO? The balanced equation is 4NH3 + 6NO --&g

t; 5N2 + 6H2O.
Choices:

A. H2O
B. NH3
C. NO
D. N2

1 answer:

givi [52]2 years ago

5 0

The limiting reactant in the reaction is the NO molecule.

<h3>What is limiting reactant?</h3>

The limiting reactant is the reactant that is present in the least amount in the reaction. The rate of reaction depends on the limiting reactant.

Given the reaction; 4NH3 + 6NO --> 5N2 + 6H2O.

Number of moles of NH3 = 43.4 g/17 g/mol = 2.55 moles

Number of moles of NO = 30 g/30 g/mol = 1 mole

Now if 4 moles of NH3 reacts with 6 moles of NO

2.55 moles of NH3 reacts with 2.55 moles * 6 moles/ 4 moles

= 3.8 moles

Hence, the limiting reactant in this reaction is NO.

Learn more about limiting reactant:brainly.com/question/14225536

#SPJ1

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A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in

mojhsa [17]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of chemist's working solution is $5.90\times 10^{-4}M$

<u>Explanation:</u>

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock magnesium fluoride solution

$M_2\text{ and }V_2$ are the molarity and volume of chemist's magnesium fluoride solution

We are given:

$M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL$

Putting values in above equation, we get:

$0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M$

Hence, the concentration of chemist's working solution is $5.90\times 10^{-4}M$

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3 years ago

A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo

Ksenya-84 [330]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$80 = x*20$

$80 = 20\:x$

$20\:x = 80$

$x = \dfrac{80}{20}$

$\boxed{x = 4}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{43}{2^{4}}$

$m = \dfrac{43}{16}$

$\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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4 years ago

A gardener cut her finger while pruning shrubs. She noticed that after two weeks the wounds had nearly healed with just a small

kkurt [141]

Answer:

c

Explanation:

7 0

3 years ago

Read 2 more answers

How much water, in grams, can be made from

Aleks [24]

Answer:

$m_{H_2O}=73.0gH_2O$

Explanation:

Hello there!

In this case, since the formation of water from hydrogen and oxygen is:

$2H_2+O_2\rightarrow 2H_2O$

Whereas we find a 2:2 mole ratio of hydrogen to water. In such a way, by using the Avogadro's number, the aforementioned mole ratio and the molar mass of water (18.02 g/mol), we obtain the following grams of water product:

$m_{H_2O}=2.44x10^{24}molec*\frac{1molH_2}{6.022x10^{23}molec}*\frac{2molH_2O}{2molH_2}*\frac{18.02gH_2O}{1molH_2O}\\\\ m_{H_2O}=73.0gH_2O$

Regards!

6 0

3 years ago

In a coffee-cup calorimeter, 150.0 mL of 0.50 M HCl is added to 50.0 mL of 1.00 M NaOH to make 200.0 g solution at an initial te

Zigmanuir [339]

Answer:

51.54°C the final temperature of the calorimeter contents.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol$

$moles=Molarity\times Volume (L)$

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

$n=0.50 M\times 0.150 L=0.075 mol$

Molarity of NaOH= 1.00 M

Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

$n'=1.00 M\times 0.050 L=0.050 mol$

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

$-56 kJ/mol=-\frac{Q}{n}$

$Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J$

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

$q=mc\times (T_{final}-T_{initial})$

where,

q = heat gained = 2800 J

c = specific heat of solution = $4.184 J/^oC$

$T_{final}$ = final temperature = $?$

$T_{initial}$ = initial temperature = $48.2^oC$

Now put all the given values in the above formula, we get:

$2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC$

$T_{final}= 51.54^oC$

51.54°C the final temperature of the calorimeter contents.

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3 years ago