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2 years ago

What is the concentration of a solution of HBr if 0.40 L is neutralized by 0.20 L of 0.50M solution of LiOH?

1 answer:

4 0

Since the mole ratio is 1 to 1 for this reaction, you can just use M1V1 = M2V2 to solve, given that 1 refers to HBr and 2 refers to LiOH.

M1 = ?
V1 = 0.40 L
M2 = 0.50 M
V2 = 0.20 L

M1(0.40) = (0.50)(0.20)

Solve for M1 —> M1 = (0.50)(0.20)/0.40 = 0.25 M HBr

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The temperature of 15.71 grams of gold rises from 32°C to 1,064°C, and then the gold melts completely. If gold’s specific heat i

antoniya [11.8K]

Answer:Total energy gained by 15.71 g is 3090.6471 joules

Explanation:

Given:

Q = heat gained by the 15.71 gram mass of gold

$Q=mc\Delta T +m\Delta H_{fusion}$

$\Delta T=(T_{final}-T_{initial})=(1064^oC-32^oC)=1032^oC$

c = specific heat capacity of gold = 0.1291 $joules/gram^oC$

m = mass of gold =15.71 g

$Q=15.71 g\times 0.1291 joules/gram^oC (1032^oC) +15.71 g\times 63.5 Joules/gram$

Heat gained by gold = $2093.0621 Joules + 997.585 Joules$

=3090.6471 joules

6 0

2 years ago

Do all atoms only have one oxidation state

MA_775_DIABLO [31]

I believe so! If not tell me So I can delete this

4 0

2 years ago

Read 2 more answers

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

2 years ago

Help please chemistry question

Gemiola [76]

Answer:

D. $K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}$

Explanation:

The general form of an equilibrium constant expression is

$K = \frac{[\text{Products}]}{[\text{Reactants}]}$

In the equilibrium

HNO₂ ⇌ H⁺ + NO₂⁻

The products are H⁺ and NO₂⁻, and the reactant is HNO₂.

∴ $K_{a} = \frac{[\text{H}^{+}][\text{NO}_{2}^{-}]}{[\text{HNO}_{2}]}$

3 0

3 years ago

When aluminum, Al, metal is dipped in an aqueous solution of hydrochloric acid, HCl, hydrogen gas, H2, is produced with the form

Nataliya [291]

Answer:

Al(s) + 3HCl(aq) → AlCl₃(aq) + 3/2H₂(g).

Explanation:

We should apply the law of conversation of mass to balance any chemical reaction; that the no. of atoms in the reactants side is equal to that in the products side.

So, the balanced equation is:

7 0

2 years ago

Read 2 more answers