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3 years ago

Webb has calculated the percent composition of a compound. How can he check his result?

Chemistry

2 answers:

wel3 years ago

8 0

Answer:

The answer is B

Explanation:

goldenfox [79]3 years ago

7 0

by making sure they are in the lowest ratio. by adding them to see if they total 100. by checking that they are whole-number multiples. by dividing them by the molar mass.

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PLEASE HELP!!!! i give a lot of points

hammer [34]

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5 0

2 years ago

Consider the following reaction where Kc = 1.80×10-2 at 698 K:

Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

$Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}$

In this case:

$Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}$

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

$[H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}$ =2.09*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}$ =4.14*10⁻² $\frac{moles}{liter}$

$[I_{2} ]=\frac{0.280 moles}{1 Liter}$ = 0.280 $\frac{moles}{liter}$

So,

$Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }$

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

8 0

3 years ago

A pulse with an amplitude of 3+ would be considered:

alexdok [17]

A pulse with an amplitude of 3+ would be considered as increased. Peripheral Pulse Assessment Grading System is measured in 0 - 3 Scale. 0 = absent 1+ = Weak/thready pulse 2+ Normal Pulse 3+ = Full, firm pulse. from the above scale we can find that the 3+ reading shows that the pulse is increased.

4 0

3 years ago

Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v

BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

For the balanced reaction:

2P₂ + 5O₂ ⇄ 2P₂O₅.

It is clear that 2 mol of P₂ react with 5 mol of O₂ to produce 2 mol of P₂O₅.

Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

Using cross multiplication:

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

Finally, we can get the volume of oxygen using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

3 0

3 years ago

3- AR 385-63/MCO 3570.1C is used in conjunction with

Rasek [7]

AR 385-63/MCO 3570.1C is used in conjunction with DA PAM 385-63.
AR 385-63/MCO 3570.1C is the regulation or order which provides revised range safety policy for the Army and Marine Corps.
This order/regulation applies on the Active Army or the Army National Guard of the United states.

8 0

3 years ago