The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).
Answer
5
Explanation:
We can go about this using the percentage compositions.
First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.
0.96/1.5 * 100 = 64%
Hence the percentage by mass of the water present is 36%
The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol
The molar mass of the water is 2(1) + 16 = 18g/mol
Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles
The total mass of the copper sulphate hydrate is 160+ 18x
Now how do we get x? Like it is said earlier, the percentage composition is constant.
Hence, 64/100 * (160 + 18x) = 160
16000 = 64(160 + 18x)
16000 = 10,240 + 1152x
16,000 - 10,240 = 1152x
1152x = 5760
x = 5760/1152
x = 5
Answer:
Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:
Thus, we solve for the final pressure P2 to obtain it as shown below:
Hence, we notice that the temperature doubles as well as the pressure.
Best regards.
So here's how you find the answer:
Given: (rate constants)
K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.
Use the equation:
ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
Transpose:
Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)
Substitute within the given transposed equation:
<span>Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).
</span>
Continuing the solution we get:
<span>Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)
</span>
The value of EA is:
<span>Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.</span>
Answer
its a potassium elements
Explanation:
K 2,8,8,1 period num 4
group 1A
or 1s1 ,2s2 2p6, 3s1
its a metal reacted rapidly with water to form a colorless basic solution of potassium hydroxide (KOH) and hydrogen gas (H2). The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.
