Answer:
2Fe + 6HC2H3O2 → 2Fe(C2H3O2)3 + 3H2
Explanation:
There you go
Answer:
16.5moles of CO2
Explanation:
Molar mass of O2 is 32g/Mol
No of mole = 880/32 = 27.5 moles
27.5 Mol of O2 gives X mole of CO2
5 mole of O2 gives 3 moles of CO2
No of mole of CO2 = (27.5×3)/ 5 = 16.5moles
Answer is: d) Hg.
Mercury is a chemical element with the symbol Hg and atomic number 80. <span> Mercury is the only metallic element that is liquid at standard conditions for temperature and pressure.
</span>Absolute viscosity of mercury is 0,0015 Pa·s.
The viscosity<span> of a </span>fluid<span> is a measure of its </span>resistance<span> to gradual deformation by </span>shear stress<span> or </span><span>tensile stress</span>
Answer:
1.46mol of Al2(SO4)3 weights 499.5 grams
Explanation:
First we will find the gram formula mass of Al2(SO4)3.
Al * 2 = 26.98 g/mol * 2 = 53.96 g
SO4 * 3 = 96.06 g/mol * 3 = 288.18
53.96 + 288.18 = 342.14 g
That is the mass of 1 mol of AL2(SO4)3.
Now we will multiply 342.14g by 1.46.
342.14 * 1.46 = 499.5244 g
Or ... 499.5 g
Answer:
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Explanation:
Step 1: Data given
iron = Fe = solid = Fe(s)
chlorine = Cl2 = gas = Cl2(g)
iron(III) chloride = FeCl3 = solid = FeCl3(s)
Step 2: The unbalanced equation
Fe(s) + Cl2(g) → FeCl3(s)
Step 3: Balancing the equation
Fe(s) + Cl2(g) → FeCl3(s)
On the left we have 2x Cl (in Cl2) and on the right side we have 3x Cl (in FeCl3). To balance the amount of Cl we have to multiply Cl2 (on the left) by 3 and FeCl3 by 2.
Fe(s) + 3Cl2(g) → 2FeCl3(s)
On the left side we have 1x Fe and on the right side we have 2x Fe (in 2FeCl3). To balance the amount of Fe, we have to multiply Fe on the left side by 2. Now the equation is balanced.
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
