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8 days ago

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A sample of a compound that contains only the elements C, H, and N is completely burned in O2 to produce 44.0 g of CO2, 45.0 g o

f H2O, and some NO2 . A possible empirical formula of the compound is

1 answer:

alisha [964]8 days ago

3 0

Clarification:

To obtain the specific element, you should multiply the grams provided by the ratio of grams of that particular element within its complete compound.

Since the query did not indicate the amount of NO2 produced, we can consider its mass to be negligible, thus assigning 1 mole to Nitrogen.

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Why is it important to have regular supervision of the weight and measurements in the market

lions [1003]

Answer:

Oversight of weights and measures ensures correct evaluations of goods and services so that everyone receives a fair exchange in the marketplace. It also acts as a deterrent, promoting honesty among traders.

Explanation:

4 0

15 days ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Tems11 [854]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0

16 days ago

In how many grams of water should 25.31 g of potassium nitrate (kno3) be dissolved to prepare a 0.1982 m solution?

lions [1003]

Solution:

Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.

Thus,

Molality = moles of solute / kg of solvent.

Therefore, kg of solvent = moles of solute / molality.

moles of solute = mass / molar mass

= 25.31 g / 101.1 g/mole

= 0.2503 mole.

kg of solvent = 0.2503 mole / 0.1982 m

= 1.263 kg

= 1263 g.

This is the final answer.

6 0

5 days ago

Read 2 more answers

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

eduard [944]

Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

Rationale:

Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

8 0

2 days ago

(a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compo

castortr0y [927]

Answer:

La masa molar del compuesto es: 168.82 g/mol

La masa molar del gas es: 16.38 g/mol

Explanation:

(a)

Utilizando la ecuación de gases ideales:

$PV=nRT$

donde,

P es la presión

V es el volumen

n es el número de moles

T es la temperatura

R es la constante de los gases, cuyo valor es = 0.0821 L.atm/K.mol

Además,

Moles = masa (m) / Masa molar (M)

La densidad (d) = Masa (m) / Volumen (V)

Así, la ecuación de gases ideales se puede expresar como:

$PM=dRt$

Dado que:-

Presión = 20 kPa = 20000 Pa

La expresión para la conversión de presión en Pascal a presión en atm se muestra a continuación:

P (Pa) = $\frac {1}{101325}$ P (atm)

20000 Pa = $\frac {20000}{101325}$ atm

Presión = 0.1974 atm

Temperatura = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Masa molar =?

Aplica la fórmula:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

La masa molar del compuesto es: 168.82 g/mol

(b)

Dado que:

Presión = 152 Torr

Temperatura = 298 K

Volumen = 250 cm³ = 0.25 L

Utilizando la ecuación de gases ideales:

$PV=nRT$

R = $62.3637\text{torr}mol^{-1}K^{-1}$

Aplicando la fórmula:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Dado que:

Masa del gas = 33.5 mg = 0.0335 g

Masa molar =?

La fórmula para calcular los moles se muestra a continuación:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Así,

$0.002045\ moles
= \frac{0.0335\ g}{Molar\ mass}$

La masa molar del gas es: 16.38 g/mol

5 0

1 day ago