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ch4aika [34]
2 years ago
3

Gaseous ammonia chemically reacts with oxygen (O2)gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of a

mmonia needed to
produce 0.090 mol of water.
Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
romanna [79]2 years ago
4 0
0.072 moles of water () produced by the reaction of 0.060mol of oxygen.
Explanation : Moles of oxygen() given = 0.060 mole
Moles of water () we need to calculate.
Step 1 : Write the balanced chemical equation.

According to balanced equation 5 mole of reacts with 4 mole of to give 4 moles of NO and 6 moles of .
Step 2 : Calculate moles of water () using moles of using mole ratio.
A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio is determined by examining the coefficients in front of formulas in a balanced chemical equation.
The coefficient of is 6 and coefficient of is 5, So mole ratio of to is 6 : 5
Moles of =


0.072 moles of is produced by the reaction of 0.060 mole oxygen ()
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Witch of the following isotopes would most likely be unstable and therefore radioactive?
STALIN [3.7K]
The answer is D; Mercury-194
All of the others are not when I looked them up

5 0
2 years ago
what is the new pressure acting on a 2.5 l balloon if its original volume was 5.8 l at 3.7 ATM of pressure
Ivan

Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 3.7 atm

P_2 = final pressure = ?

V_1 = initial volume = 5.8 L

V_2 = final volume = 2.5 L

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

3.7atm\times 5.8L=P_2\times 2.5L

P_2=8.6atm

Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.

8 0
2 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
Jamal wants to make a model of a hill near his house to test the way the slope affects how rain runs down the hill. Which type o
Nikitich [7]

a scale-model mound made of the same materials that make the real hill

4 0
2 years ago
Read 2 more answers
A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?
REY [17]

Answer:

Sn_2O

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1

Therefore, the empirical formula is:

Sn_2O

Best regards.

8 0
2 years ago
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