Answer:
1.1724 g.
Explanation:
- Firstly we need to calculate the percentage of pure MgSO₄ in (MgSO₄.7H₂O).
- If we have 1.0 mol of MgSO₄.7H₂O, then we will have the the mass of its molecular mass.
The molecular mass of 1.0 mol (MgSO₄.7H₂O) = 246.4 g/mol.
The molecular mass of pure MgSO₄ = 120.366 g/mol.
The molecular mass of 7(H₂O) = 7(18.0 g/mol) = 126.0 g/mol.
∴ The mass % of pure MgSO₄ = [(mass of pure MgSO₄)/(mass of MgSO₄.7H₂O)] x 100 = [(120.366 g/mol)/(246.4 g/mol)] x 100 = 48.85%.
∴ the quantity of pure MgSO₄ = (mass of MgSO₄.7H₂O)(% of MgSO₄/100) = (2.4 g)(48.85/100) = 1.1724 g.
E kinetic enerhy of the air particles will increaase
2 H₂ + O₂ → 2 H₂O
Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O
CH₄ + 2 O₂ → CO₂ + 2 H₂O
Explanation:
To balance the chemical equations the number and type of atoms entering the reaction have to be equal to the number an type of atoms leaving the reactions. To do this we add coefficients (usually integer numbers) before each substance.
So we have the following balanced chemical reactions:
2 H₂ + O₂ → 2 H₂O
Al₂O₃ + 6 HCl → 2 AlCl₃ + 3 H₂O
CH₄ + 2 O₂ → CO₂ + 2 H₂O
Learn more about:
balancing chemical equations
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Answer:
I think it would be a,b,c,d,e but im not sure
Explanation:
Can I get brainlest plz?
Answer:
76%
Explanation:
The lab grade of the student is made of report sheets, lab exams and lab quizzes with different weights for each score. To find the lab grade we need to multiply each score with respective weight. He has 86% in the report sheet which has 35% weight, 66% in the lab exam which have 35% weight, and 76% in the quizzes which have 30% weight. The score will be:
86%*35% + 66%*35% + 76%*30%= 76%
His score for the lab grade will be 76%