Answer:
3.39g of MgBr
Explanation:
Reaction between magnesium and bromine
Mg + Br₂ → MgBr
Mass of MgBr = ?
1g of Mg + 5.0g of Br₂
6g of Mg and Br₂
Molar mass of Mg = 24g/mol
Molar mass of Br = 80g/mol
Number of moles = mass / molar mass
Mass = number of moles * molarmass
Mass of Mg = 1 * 24 = 24g
Mass of Br₂ = 1 * (2*80) = 160g (diatomic molecule)
Molarmass of MgBr = (24 + 80) = 104g/mol
Mass of MgBr = 1 * 104 = 104g/mol
24g of Mg + 160g of Br₂ = 104g of MgBr
184g of (mg and Br) = 104g of mgBr
6g of (mg and Br) = y g of MgBr
y = (6 * 104) / 184
y = 3.39g of MgBr
Answer: The mass of blue copper sulfate is 3.5 g
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the heating of copper sulfate crystals is:
Let the mass of blue copper sulfate be 'x' grams
We are given:
Mass of copper sulfate powder = 2.1 grams
Mass of water = 1.4 grams
Total mass on reactant side = x
Total mass on product side = (2.1 + 1.4) g
So, by applying law of conservation of mass, we get:
Hence, the mass of blue copper sulfate is 3.5 grams
Answer:Subatomic particle Symbol Relative charge
Proton p +1
Neutron n 0
Electron e- -1
Explanation:
These are the names, symbols, and charges. Hope I helped!
Answer:
T₂ = 51826.1 K
Explanation:
Given data:
Initial Volume = 2.3 L
Final volume = 400 L
Initial temperature = 25 °C (25+ 273 = 298 K)
Final temperature = ?
Solution:
V₁/T₁ = V₂/T₂
T₂ = V₂ T₁/V₁
T₂ = 400 L . 298 K / 2.3 L
T₂ = 119200 K. L / 2.3 L
T₂ = 51826.1 K
Answer:I dont know the answer but i need the points thx!!!!
Explanation: