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3 years ago

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A 100.-gram sample of H2O(l) at 22.0°C absorbs 830. joules of heat. What will be the final temperature of the water? Hurry pleas

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1 answer:

k0ka [10]3 years ago

4 0

Answer:

That is extremely confusing. Try contacting your prof.

Explanation:

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How are the sizes of stars measured?<br><br><br> I just need a simple answer

S_A_V [24]

They are determined by the length of time it takes for light to travel from the Earth to the desired star

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3 years ago

On the pH scale, a 7 indicates _____.<br> a base<br> an acid<br> neutrality

worty [1.4K]

The correct answer is:

neutrality

on the pH scale 7 is neutral above 7 is basic and below 7 is acidic.

Explanation:

The pH is an example of how acidic/basic water is. The scale goes from 0 - 14, with 7 being neutral, pH of less than 7 designates acidity, whereas a pH of greater than 7 intimates a base. pH is really a pattern of the relative amount of free hydrogen and hydroxyl ions in the water.pH is a measure of how acidic/basic water is.

It is neutral because, at this pH, the molar concentrations of Hydroxide and Hydronium are same, therefore the substance is not a base nor an acid.

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3 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

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it's d. H:C:C:F

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