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3 years ago

If 2.03 g of oxygen react with carbon monoxide, how many grams of CO2 are formed?

Chemistry

1 answer:

mamaluj [8]3 years ago

8 0

Answer:

The suitable equation for this reaction is

2CO + O₂ -----> 2CO₂

Here, we are given that we have 2 grams of O₂

From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂

Moles of O₂:

2/32 = 1/16 moles

Therefore, the number of moles of CO₂ is twice the moles of O₂

Moles of CO₂ = 2 * 1/16

Moles of CO₂ formed = 1/8 moles

Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂

Mass of CO₂ formed = 44 * 1/8

Mass of CO₂ formed = 5.5 grams

Hence, option B is correct

Kindly Mark Brainliest, Thanks!!!

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Explain what might happen to tundra animals, such as polar bears, as earth's climate warms. pls explain

SCORPION-xisa [38]

Answer:

They will die.

Explanation:

Polar Bears were made to live in the frigid temperatures of the tundras. If there is a temperature change it can affect their habitat. It can destroy it and they would have to move out to another place. They can die due to global warming.

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3 years ago

In at least three sentences, explain three disadvantages associated with hydrogen fuel use.

Dmitry_Shevchenko [17]

Here are some disadvantages, is that nitrogen dioxide is a toxic gas and it can still be harmful when ingested by human, also critics of hydrogen fuel cells argue that although these cells do not emit carbon after burning, they give out nitrogen dioxide and other emissions.

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3 years ago

Your dad is working on creating a brick border for the lake in your backyard each brick has a mass of 100 g and a volume of 20 c

wel

Answer:

5000kg/m³

Explanation:

density=mass/volume

d=m/v

d=100/20

=5g/cm³

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5g/m³*1000=5000kg/m³

4 0

3 years ago

Read 2 more answers

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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3 years ago

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3 years ago