First, we calculate of the concentration of the H+ ions in the solution from the pH given. Then, calculate the new concentration after dilution. Calculation are as follows:
pH = -log[H+]
5 = -log[H+]
[H+] = 1 x 10^-5 M
M1V1 = M2V2
<span>1 x 10^-5 M (V1) = M2(100V1)
</span>M2 = 1 x 10^-7
pH = -log[<span>1 x 10^-7</span>]
pH = 7
Answer:
4. Atmospheric pressure decreases as altitude increases.
Explanation:
hope this helps and is right. p.s i really need brainliest :)
Answer:
Explanation:
Ksp(BaSO4)=1.07×10−10
BaSO₄ → Ba²⁺ + SO₄²⁻
1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)
but Ba²⁺ = 1.3×10⁻² M
1.07×10⁻¹⁰ = 1.3×10⁻² M × ( SO₄²⁻)
( SO₄²⁻) = 1.07×10⁻¹⁰ / 1.3×10⁻² = 0.823 × 10⁻⁸ M
while Ksp(CaSO4)=7.10×10−5
CaSO₄ → Ca²⁺ + SO₄²⁻
7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)
( SO₄²⁻) = 7.10×10⁻⁵ / 2.0×10⁻² = 3.55 × 10⁻³ M
comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄
The element has 83 electrons
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
