Answer:
The element is strontium and the number of neutrons it have is 51.
Explanation:
Based on the given information, the ionic compound is,
XCl₂ ⇔ X₂⁺ + 2Cl⁻
X2+ is the ion of the mentioned element
As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,
No. of electrons = atomic number - charge
36 = atomic number - 2
Atomic number = 38
Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.
Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,
No. of neutrons = mass number - atomic number
= 89 - 38
= 51 neutrons.
Answer:
6.53g of K₂SO₄
Explanation:
Formula of the compound is K₂SO₄
Given parameters:
Volume of K₂SO₄ = 250mL = 250 x 10⁻³L
= 0.25L
Concentration of K₂SO₄ = 0.15M or 0. 15mol/L
Unknown:
Mass of K₂SO₄ =?
Methods:
We use the mole concept to solve this kind of problem.
>>First, we find the number of moles using the expression below:
Number of moles= concentration x volume
Solving for number of moles:
Number of moles = 0.25 x 01.5
= 0.0375mole
>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:
Mass(g) = number of moles x molar mass
Solving:
To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.
For:
K = 39g
S = 32g
O = 16g
Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)
= 78 +32 + 64
= 174g/mol
Using the expression:
Mass(g) = number of moles x molar mass
Mass of K₂SO₄ = 0.0375 x 174 = 6.53g
Answer:homogeneous mixture
A sandstorm is a homogeneous mixture.
Explanation:
Answer:
0.51
Explanation:
Given the Nernst equation;
E= E° - 0.0592/n logQ
E= 355 mV or 0.355 V
E° = 0.34 - 0= 0.34 V
n= 2(two electrons were transferred in the process)
Equation of the reaction;
H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)
Substituting values;
0.355 = 0.34 - 0.0592/2 log([H^+]/1)
0.355 - 0.34 = - 0.0296 log [H^+]
0.015/-0.0296 = log [H^+]
Antilog (-0.5068) = [H^+]
[H^+] = 0.311 M
pH = -log[H^+]
pH= - log(0.311 M)
pH = 0.51
