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3 years ago

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Chemistry

1 answer:

NARA [144]3 years ago

6 0

Answer:

K⁺ and HPO₄²⁻ ions are present in K₂HPO₄

Explanation:

K₂HPO₄ consist of anion and cation such as HPO₄²⁻ and K⁺. We can see that the charge on HPO₄²⁻ is negative 2. Thus inroder to balance the charge and to neutral the compound two potassium ions are attached because the charge on one atom of potassium is K⁺.

Properties of K₂HPO₄ :

It is salt of phosphoric acid.

Its density is 2.44 g/cm³.

It is odourless compound.

It is white powder and soluble in water.

It is used in fertilizer because it provide phosphorus which is beneficial for the growth of plants.

It is also used as a additive in food.

It is inorganic compound and also used as buffering agent.

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The cyanide ion (CN-1) is highly toxic, but can be remediated in solution by redox reaction with the hypochlorite ion (the princ

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Answer:

2.275x10⁶g of CN⁻¹ were dissolved in the sample

Explanation:

Molarity is defined as the ratio between moles of solute and liters of solution.

If a CN⁻¹ solution has a concentration of 25mM, there are 0.025 moles of CN⁻¹ per liter of solution.

If the sample has a colume of 3.5x10⁶L, moles of CN⁻¹ are:

3.5x10⁶L × (0.025moles / L) = <em>8.75x10⁴ moles of CN⁻¹ in the sample of water.</em>

In grams (As molar mass of CN⁻¹ is 26g/mol):

8.75x10⁴ moles CN⁻¹ × (26g / mol) = <em>2.275x10⁶g of CN⁻¹ were dissolved in the sample</em>

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Answer:

The mean is $x=0.503\frac{mg}{L}$

The 90% confidence interval is:

$i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]$

Explanation:

1. First organize the data:

$x_{1}=0.487$

$x_{2}=0.487$

$x_{3}=0.511$

$x_{4}=0.511$

$x_{5}=0.519$

As there are 5 data, the sample size (n) is n=5

2. Calculate the mean x:

The mean is calculated adding up all the data and divide them between the sample size.

$x=\frac{0.511+0.487+0.511+0.487+0.519}{5}$

$x=0.503\frac{mg}{L}$

3. Find 90% confidence interval.

The formula to find the confidence interval is:

$i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})]$ (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

$1-\alpha=0.90$

$\alpha=0.10$

$\frac{\alpha}{2}=0.05$

$z_{0.05}=1.645$

4. Find the standard deviation

$d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}$

$d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}$

$d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}$

$d=\sqrt{2.24*10^{-4}}$

$d=0.015$

5. Replace values in (Eq.1):

$i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]$

For the addition:

$i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.514$

For the subtraction:

$i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.492$

The 90% confidence interval is:

$i_{0.90}=[0.492,0.514]$

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