Which of the following is the smallest volume? 500 mL 2,500 cm3 5.5 x 10-1 L 25 m3

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

3 years ago

A train travels 65 km in three hours and then 52 km in two hours. What is its average speed?

Darya [45]

the average speed is 23.4 km per hour cause it is total distance over total time equal to average speed

5 0

3 years ago

This cold river of air travels from west to east in the northern hemisphere. In winter, it dips south helping to bring cooler ai

Alex_Xolod [135]

Answer:

Polar Jet stream

Explanation:

Polar Jet stream is also called the polar front jet or mid latitude jet steam, it is a very powerful belt of the upper level winds which sits above the polar front. It is the strongest wind in the tropopause, it's movement is towards the westerly direction of the mid latitude.

4 0

3 years ago

How much heat is required to warm 1.50L of water from 25.0C to 100.0C? (Assume a density of 1.0g/mL for the water.)

Masteriza [31]

<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{1500mL}\\\\\text{Mass of water}=(1g/mL\times 1500mL)=1500g$

To calculate the heat absorbed by the water, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 1500 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$q=1500g\times 4.186J/g^oC\times 75^oC=470925J=470.9kJ$

Hence, the amount of heat required to warm given amount of water is 470.9 kJ

6 0

3 years ago

Please help.

mojhsa [17]

Explanation: An element is represented in the form of $_{A}^{Z}\textrm{X}$ where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

and Atomic Mass = Number of neutrons + Number of protons

Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

The nearest noble gas to the element having atomic number 120 is Oganesson (Og), which has an atomic number of 118, so the next two electrons will be filled in the 8s orbital.

Electronic Configuration of Jw is $[Og]8s^2$

This electronic configuration lets us know about the location of the element in periodic table.

As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

8 0

3 years ago