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3 years ago

What’s the balanced equation for the reaction between zinc oxide and dilute hydrochloric acid ?

Chemistry

1 answer:

larisa [96]3 years ago

7 0

Answer:

ZnO + 2HCl → ZnCl₂ + H₂O

Explanation:

The unbalanced reaction between zinc oxide (ZnO) and dilute hydrochloric acid (HCl) is:

ZnO + HCl → ZnCl₂ + H₂O

Now we'll procede to balance the equation:

There is 1 Cl atom on the left side and 2 on the right, so we change that:

ZnO + 2HCl → ZnCl₂ + H₂O

The number of atoms for each element is the same on both sides, so the equation is now balanced.

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'

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Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

$\displaystyle M_1V_1= M_2V_2$

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

$\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}$

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
$\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}$

Convert this value to mL:
$\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}$

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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