Answer: 899 g
Explanation:
Since the mass of a mole of diatomic chlorine is about 70.9 g/mol, this means that 875 grams is about 875/70.9 = 12.3 mol.
Hence, we know that chlorine is the limiting reactant, and that about 12.3 moles of chlorine will be consumed.
This means that 12.3(2)=24.6 grams of HCl will be produced, which will have a mass of about (24.6)(36.46)=899 g
Answer:
Sodium Hydroxide + Sulfuric Acid = Sodium Sulfate + Water
2NaOH + H2SO4 → Na2SO4 + 2H2O
Explanation:
To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above.
Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.
Ionic charges are not yet supported and will be ignored.
Replace immutable groups in compounds to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will.
Compound states [like (s) (aq) or (g)] are not required.
You can use parenthesis () or brackets [].
Answer:
101.63° C
Explanation:
Volume expansivity γa = γr - γ g = 18 × 10⁻⁵ - 2.0 × 10⁻⁵ = 16 × 10⁻⁵ /K
v₂ - v₁ / v₁θ = 16 × 10⁻⁵ /K
(500 - 492 ) mL / (492 × 16 × 10⁻⁵) = θ
θ = 101.63° C
Answer:

Explanation:
Due to Coulomb´s law electric force can be described by the formula
, where K is the Coulomb´s constant (
),
= Charge 1 (Na+ in this case),
is the charge 2 (Cl-) and r is the distance between both charges.
Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is
.
so we have ![W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]](https://tex.z-dn.net/?f=W%3DW_%7Bf%7D%20-W_%7Bi%7D%20%3D%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Drf%7D%7Br_%7Bf%7D%20%5E%7B2%7D%7D%29-%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Dri%7D%7Br_%7Bi%7D%20%5E%7B2%7D%7D%29%3DKq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%5B%5Cfrac%7B1%7D%7B%7Br_%7Bf%7D%7D%7D%20-%5Cfrac%7B1%7D%7B%7Br_%7Bi%7D%7D%7D%5D)
Given that ri= 1.1nm=
and rf= infinite distance
![W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J](https://tex.z-dn.net/?f=W%3D%289x10%5E%7B9%7D%29%281.6x10%5E%7B-19%7D%29%28-1.6x10%5E%7B-19%7D%29%5B%5Cfrac%7B1%7D%7B%5Calpha%20%7D-%5Cfrac%7B1%7D%7B%281.1x10%5E%7B-9%7D%29%7D%5D%3D2.1x10%5E%7B-19%7DJ)
Answer:
B
Explanation:
found it from a person wjwjdjjanqnqq