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2 years ago

How many metres further would the sound intensity be 1/2 times as much as that at a distance of 3 m from an acoustic source?

Chemistry

1 answer:

marshall27 [118]2 years ago

7 0

Solution :

Let, intensity at distance 3 m from the acoustic source is $I_o$ .

We need to find the distance d where the intensity is $\dfrac{I_o}{2}$ .

Now, we know intensity is inversely proportional to square of distance between the source.

$\dfrac{I_o}{\dfrac{I_o}{2}} = \dfrac{d^2}{3^2}\\\\d^2 = 18 \\\\d = 3\sqrt{2}\ m$

Therefore, intensity will be halved at a distance of $3\sqrt{2}\ m$ .

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3 years ago

What is the best way to reduce water pollution

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2 years ago

Read 2 more answers

3. What might happen to the ecosystem if this population were removed?<br><br> please help

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I don't know the population, but I know that if one population is removed then the population that consumes it will die off, and the population that the removed one consumes will grow out of control.

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3 years ago

How many grams of ba(io3)2 can be dissolved in 250. Ml of distilled water at 25 oc ? Note: ksp of ba(io3)2 is 1.57 × 10-9 and mo

N76 [4]

$0.142 \; \text{g} \; \text{Ba}(\text{IO}_3)_2$

Explanation

Solid barium iodate and its ionic components are in a dynamic equilibrium when dissolved in an aqueous solution. Each mole of barium iodate $\text{Ba}(\text{IO}_3)_2$ dissolves to produce one mole of barium $\text{Ba}^{2+}$ ions and two moles of iodate $\text{IO_3}^{-}$ ions.

$\text{Ba}(\text{IO}_3)_2 \; (aq) \leftrightharpoons \text{Ba}^{2+} \; (aq) + 2 \; \text{IO}_3^{-} \; (aq)$

Thus

$K_{\text{sp}} &= & [\text{Ba}^{2+}] \cdot [\text{IO}_3^{-}]^{2}$ .

Given that

$K_{\text{sp}}$ of $1.57 \times 10^{-9}$ from the question, and

$[\text{Ba}(\text{IO}_3)_2, \; \text{dissolved}] = [\text{Ba}^{2+}] = 2 \; [\text{IO}_3^{-}]$ as seen in the dissociation equilibrium,

$\begin{array}{lll}[\text{Ba}(\text{IO}_3)_2, \; \text{dissolved}]^{3} &=& [\text{Ba}^{2+}] \cdot [\text{IO}_3^{-}]^{2} \\ & = & K_{sp}\\&=&1.57\times 10^{-9} \end{array}$

$[\text{Ba}(\text{IO}_3)_2, \; \text{dissolved}] = \sqrt[3]{K_{\text{sp}}} = 1.16 \times 10^{-3} \; \text{mol}\cdot \text{dm}^{-3}$

$250 \; \text{ml} = 0.250 \text{dm}^{-3}$ of distilled water can thus dissolve up to

$\begin{array}{lll} n &=& c \cdot V\\ &= & 1.16 \times 10^{-3} \; \text{mol} \; \text{Ba}(\text{IO}_3)_2\cdot \text{dm}^{-3} \times 0.250 \; \text{dm}^{-3} \\ & = & 2.91 \times 10^{-4} \; \text{mol} \; \text{Ba}(\text{IO}_3)_2\; \text{,}\end{array}$

which corresponds to

$\begin{array}{lll} m &= & n \cdot M\\ &= & 2.91 \times 10^{-4} \; \text{mol} \; \text{Ba}(\text{IO}_3)_2 \times 487.1323 \; \text{g} \cdot \text{mol}^{-1} \\ & = &0.142 \; \text{g} \; \text{Ba}(\text{IO}_3)_2 \end{array}$

8 0

3 years ago

A gas has a mass of 3.82 g and occupies a volume of 0.854 L. The temperature in the laboratory is 302 K, and the air pressure is

Rzqust [24]

The molar mass of the gas that has a mass of 3.82 g and occupies a volume of 0.854 L is 106.66g/mol.

<h3>How to calculate molar mass?</h3>

The molar mass of a substance can be calculated by dividing the mass of the substance by its number of moles.

However, the number of moles of the gas in this question needs to be calculated first using the ideal gas law equation:

PV = nRT

Where;

P = pressure
V = volume
n = number of moles
T = temperature
R = gas law constant

1.04 × 0.854 = n × 0.0821 × 302

0.888 = 24.79n

n = 0.888/24.79

n = 0.036mol

Molar mass of gas = 3.82g/0.036mol

Molar mass = 106.66g/mol

Therefore, the molar mass of the gas that has a mass of 3.82 g and occupies a volume of 0.854 L is 106.66g/mol.

Learn more about molar mass at: brainly.com/question/12127540

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2 years ago