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


0.0527 moles
Molar mass MgSO₄ = 120.4 g/mol
Mass of anhydrous magnesium sulfate :
( 0.0527 x 120.4 ) => 6.349 g
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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
Answer:
The catalyzed reaction will take 2.85 seconds to occur.
Explanation:
The activation energy of a reaction is given by:

For the reaction without catalyst we have:
(1)
And for the reaction with the catalyst:
(2)
Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

Since the reaction rate is related to the time as follow:
![k = \frac{\Delta [R]}{t}](https://tex.z-dn.net/?f=%20k%20%3D%20%5Cfrac%7B%5CDelta%20%5BR%5D%7D%7Bt%7D%20)
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}}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B2%7D%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BR%5D%2Ft_%7B1%7D%7D%7B%5CDelta%20%5BR%5D%2Ft_%7B2%7D%7D%20)


Therefore, the catalyzed reaction will take 2.85 seconds to occur.
I hope it helps you!
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
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