What is the force of an object with a mass of 12 kg and an acceleration of 4 m/s2?

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed

MaRussiya [10]

The question is incomplete, here is the complete question:

Suppose the reaction is:

$A+2B\rightarrow C$

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed

<u>Answer:</u> The maximum amount of C produced will be 1 mole

<u>Explanation:</u>

We are given:

Moles of A = 6 moles

Moles of B = 2 moles

For the given chemical reaction:

$A+2B\rightarrow C$

By Stoichiometry of the reaction:

2 moles of B reacts with 1 mole of A

So, 2 moles of B will react with = $\frac{1}{2}\times 2=1$ mole of A

As, As, given amount of A is more than the required amount. So, it is considered as an excess reagent.

Thus, B is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of B produces 1 mole of C

So, 2 moles of given B will produce $=\frac{1}{2}\times 2=1mol$ of C

Hence, the maximum amount of C produced will be 1 mole

3 0

3 years ago

How do experiments help scientists learn about the world?

arsen [322]

Answer:

A. Experiment help explain scientific theories

7 0

3 years ago

What is the mass present in a 10.0L container of oxygen at a pressure of 105kPa and 20 degrees Celsius

omeli [17]

1.31 × 10⁴ grams.

<h3>Explanation</h3>

Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:

$P \cdot V = n \cdot R\cdot T$ ,

where

$P$ the pressure on the gas, $\bf P = 10^{5}\;\textbf{kPa}=10^{8}\;\textbf{Pa}$ ;
$V$ the volume of the gas, $V = 10.0 \;\text{L} = 10.0\times 10^{-3}\;\text{m}^{3}=10^{-2}\;\text{m}^{3}$ ;
$n$ the number of moles of the gas, which needs to be found;
$T$ the absolute temperature of the gas, $T=20\;\textdegree{}\text{C} = (20 + 273.15)\;\text{K} = 293.15\;\text{K}$ .
$R$ the ideal gas constant, $R = 8.314$ if P, V, and T are in their corresponding SI units: Pa, m³, and K.

Apply the ideal gas law to find $n$ :

$n = \dfrac{P\cdot V}{R\cdot T} = \dfrac{{\bf 10^{8}\;\textbf{Pa}}\times 10^{-2}\;\text{m}^{3}}{8.314 \;\text{Pa}\cdot\text{m}^{3}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times 293.15\;\text{K}} = 410.3\;\text{mol}$ .

In other words, there are 410.3 moles of O₂ molecules in that container.

There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be ${\bf 2}\times 16.00 = 32.00\;\text{g}$ . The mass of 410.3 moles of O₂ will be:

$410.3 \times 32.00 = 1.31\times10^{4}\;\text{g}$ .

What would be the mass of oxygen in the container if the pressure is approximately the same as STP at $10^{5}\;\textbf{Pa}$ or $10^{2}\;\text{kPa}$ instead?

6 0

4 years ago

Please help me asap

KiRa [710]

A) Water vapor, carbon dioxide, methane, and ammonium

B) Plate tectonics shifted (Unsure about this)

C) Dew

D) Drinking, bathing, cooking

E) 0.04

F) Decompose.

6 0

4 years ago

El volumen de un gas en un recipiente cerrado es de 3.5 litros a una temperatura de 25oC y una atmósfera constante de presión. ¿

polet [3.4K]

Answer:

El volumen del gas en el recipiente es 3.29 L si se disminuye la temperatura a 7°C

Explanation:

La Ley de Charles es una ley de los gases que relaciona el volumen y la temperatura de una cierta cantidad de gas a presión constante.

Esta ley establece que el volumen es directamente proporcional a la temperatura del gas. Es decir que si la temperatura aumenta, el volumen del gas aumenta, mientras que si la temperatura del gas disminuye, el volumen disminuye.

Matemáticamente, si la cantidad de gas y la presión permanecen constantes, el cociente entre el volumen y la temperatura siempre tiene el mismo valor:

$\frac{V}{T}=k$

Suponiendo un cierto volumen de gas V1 que se encuentra a una temperatura T1 al comienzo del experimento, al variar el volumen de gas hasta un nuevo valor V2, entonces la temperatura cambiará a T2, y se cumplirá:

$\frac{V1}{T1}=\frac{V2}{T2}$

En este caso, sabes:

V1= 3.5 L
T1= 25 °C=298 °K (siendo 0°C=273°K)
V2=?
T2=7 °C=280 °K

Reemplazando:

$\frac{3.5 L}{298 K}=\frac{V2}{280 K}$

Y resolviendo obtienes:

$V2=280 K*\frac{3.5 L}{298 K}$

V2= 3.29 L

<u><em>El volumen del gas en el recipiente es 3.29 L si se disminuye la temperatura a 7°C</em></u>

5 0

3 years ago