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1 year ago

Nitrogen (N2) and hydrogen (H2) react to make NH3 according to the following

Chemistry

1 answer:

Schach [20]1 year ago

3 0

The number of moles of NH3 that could be made would be 0.5 moles

<h3>Stoichiometric reactions</h3>

From the balanced equation of the reaction:

N2 (g) + 3 H2(g) ----> 2NH3 (g)

The mole ratio of N2 to H2 is 1:3

Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.

Mole ratio of H2 and NH3 = 3:2

Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:

2 x 0.75/3 = 0.5 moles

More on stoichiometric calculations can be found here: brainly.com/question/8062886

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Convert 678.92 cm into meters

Anastaziya [24]

Answer:6.7892 meters

Explanation:

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2 years ago

If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50

tangare [24]

Answer:

87.5 mi/hr

Explanation:

Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.

vf = 60 mi/hr

vi = 0 mi/hr

Δt = 8 secs

a = vf - vi/ Δt

= 60 mi/hr - 0 mi/hr/ 8 secs

= 60 mi/hr / 8 secs

= 7.5 mi/hr^2

Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.

vi = 50 mi/ hr

Δt = 5 secs

a = 7.5 mi/ hr^2

a = vf - vi/ Δt

7.5 = vf - 50 mi/hr / 5 secs

37.5 = vf - 50

87.5 mi/ hr = vf

7 0

3 years ago

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of water

i = Van't Hoff factor

$K_f$ = freezing point constant

m = molality

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0

3 years ago

A sample of hydrogen occupies a volume of 351 mL at a temperature of 20 degrees Celsius. What is the new volume if the temperatu

xeze [42]

Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml

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2 years ago

Calculate the reaction quotient Qp for the following redox reaction: 14H+ + Cr2O72- + 6Cl- ----> 2Cr3+ + 3Cl2 + 7H2O The reac

stich3 [128]

Answer:

Value of $Q_{p}$ for the given redox reaction is $1.0\times 10^{-8}$

Explanation:

Redox reaction with states of species:

$14H^{+}(aq.)+Cr_{2}O_{7}^{2-}(aq.)+6Cl^{-}(aq.)\rightarrow 2Cr^{3+}(aq.)+3Cl_{2}(g)+7H_{2}O(l)$

Reaction quotient for this redox reaction:

$Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}$

Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of $H_{2}O$ is taken as 1 due to the fact that $H_{2}O$ is a pure liquid.

$pH=-log[H^{+}]$

So, $[H^{+}]=10^{-pH}$

Plug in all the given values in the equation of $Q_{p}$ :

$Q_{p}=\frac{(0.10)^{2}\times (0.010)^{3}}{(10^{-0.0})^{14}\times (1.0)\times (1.0)^{6}}=1.0\times 10^{-8}$

7 0

2 years ago