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2 years ago

Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:

Chemistry

2 answers:

fredd [130]2 years ago

8 0

The answer to your question is D. 24g

egoroff_w [7]2 years ago

6 0

The grams of oxygen that are required to produce 1 mole of H₂O is 16 g ( answer B)

calculation

2 CH₄ + 2NH₃ +3 O₂ → 2HCN + 6H₂O

step 1: use the mole ratio to find moles of O₂

from equation above the mole ratio of O₂: H₂O is 3:6 therefore the moles of O₂ = 1 mole x3/6 =0.5 moles

step 2: find mass of O₂

mass= moles x molar mass

from periodic table the molar mass of O₂ = 16 x2= 32 g/mol

mass O₂ = 0.5 moles x 32 g/mol = 16 g (answer B)

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⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

4 0

3 years ago

In this experiment, you were asked to calculate your percent yield of alum assuming that you started with pure aluminum. Actuall

neonofarm [45]

Answer:

Percent yield of alum is 38.6%

Explanation:

Find attached calculations for detailed explanation.

3 0

3 years ago

What pressure in atmospheres(atm) is equal to 45.6 kPa?

postnew [5]

1 kpa = 0.0098692327 atm so just multiply that by 45.6

4 0

3 years ago

What mass of HBO2 is produced from the combustion of 139.5 g of B2H6? Answer in units of g.

aliina [53]

Answer:

$m_{HBO_2}=441.8gHBO_2$

Explanation:

Hello there!

In this case, since the combustion of B2H6 is:

$B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O$

Thus, since there is 1:2 mole ratio between the reactant and product, the produced grams of the latter is:

$m_{HBO_2}=139.5gB_2H_6*\frac{1molB_2H_6}{27.67gB_2H_6} *\frac{2molHBO_2}{1molB_2H_6} *\frac{43.82gHBO_2}{1molHBO_2}$

$m_{HBO_2}=441.8gHBO_2$

Best regards!

5 0

2 years ago

Maya prepared 0.50 liters of a solution by dissolving 2.0 moles of an unknown compound in water. What is the molarity of the sol

Elena-2011 [213]

Answer:

4 M

Explanation:

Molarity can be represented by the following ratio:

Molarity = moles / volume (L)

Since you have been given both the mass and volume, you can plug the values into the equation and solve for molarity.

Molarity = moles / volumes

Molarity = 2.0 moles / 0.50 L

Molarity = 4 M

6 0

1 year ago