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never [62]
3 years ago
13

What molecule is this?

Chemistry
2 answers:
Sergio [31]3 years ago
6 0

Answer:

O2C

Explanation:

Mrac [35]3 years ago
6 0

Answer:

Carbon Monoxide (CO) is a colorless, odorless, tasteless, toxic gas that has the molecular formula CO. The molecule consists of a carbon atom that is triply bonded to an oxygen atom. ... The linear molecule consists of a carbon atom that is doubly bonded to two oxygen atoms, O=C=O

Explanation:

You might be interested in
If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
Paul [167]

6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

Moles =\frac{mass}{molar \;mass}

Moles =\frac{13.0 g}{246.52}

0.0527 moles

Molar mass MgSO₄ = 120.4 g/mol

Mass of anhydrous magnesium sulfate :

( 0.0527 x 120.4 ) => 6.349 g

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

4 0
1 year ago
20 mL of 80°C water is mixed with 20 mL of 0°C water in a perfect calorimeter. What is the final temperature?
dedylja [7]
To calculate for the final temperature, we need to remember that the heat rejected should be equal to the absorbed by the other system. We calculate as follows:

Q1 = Q2
(mCΔT)1 = (mCΔT)2

We can cancel m assuming the two systems are equal in mass. Also, we cancel C since they are the same system. This leaves us,

 (ΔT)1 = (ΔT)2
(T - 80) = (0 - T)
T = 40°C
8 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
The molar mass of glucose is 180.2 g/mol. How many grams of glucose will be produced when 132.0 g of CO2 reacts with an excess o
andriy [413]

Answer:

The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams

Explanation:

The chemical equation for the reaction is

6H₂O + 6CO₂  → C₆H₁₂O₆ + 6O₂

From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose  C₆H₁₂O₆ and 6 moles oxygen gas

The molar mass of CO₂ = 44.01 g/mol

There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles

However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆

and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of

mass of 1 mole C₆H₁₂O₆ = 180.2 g

mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams

Mass of glucose produced = 90.1 grams

7 0
3 years ago
HALLP ME STUCKKKKK
likoan [24]

Answer:

The average height of the sunflower sprouts at the end of week 3 is  12.0

cm.

The average height of the birch sprouts at the end of week 3 is  7.2

cm.

Explanation:

they showed it on edgu

7 0
3 years ago
Read 2 more answers
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