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3 years ago

How does a solution with a pH of 2 compare to a solution with a pH of 1?

1 answer:

5 0

I’m going to go with The pH 1 solution is two times more acidic than that with a pH of 1. If that’s not the answer, then my second best guess would be The pH solution is ten times more acidic than that with a pH of 2.

The first and third answer choice wouldn’t even be an answer so you can just cancel those two out and pick between answer choice 2 or 4.

Hope this helps, and have an amazing day!! :))

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Brainliest for an answer!

Degger [83]

The volume of H₂ : = 15.2208 L

<h3>Further explanation</h3>

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

5 0

3 years ago

How many particles would be found in a 89.2 g sample of Na?

julsineya [31]

One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert 89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles

5 0

3 years ago

While working in the lab, you observe the following volume.

MakcuM [25]

Answer:

B 18.4L

Explanation:

7 0

3 years ago

Read 2 more answers

Which of the following pairs are about equal in mass?

Mademuasel [1]

Answer:

Proton and neutrons

Explanation:

they both have a mass of 1 amu or 1 atomic mass

3 0

3 years ago

Read 2 more answers

N204(0) + 2NO2(g)

user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

$\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}$

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state

Q is the product of the ion ions from the reacting substance

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

$\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}$