Is neon a metal, nonmetal , or metalloid

bazaltina [42]

A woman and a male have sex. The man ejaculates sperm into the females uterus. If one attaches to the woman’s eggs, it will fertilize the egg and impregnate that woman. If more than one sperm attached to an egg that’s how twins are made

8 0

2 years ago

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When ammonium nitrate is added to a suspension of magnesium hydroxide in water, the Mg(OH)2 dissolves. Write a net ionic equatio

Viefleur [7K]

Answer:

Net ionic: $Mg(OH)_{2}+2NH_{4}^{+}\rightarrow Mg^{2+}+2NH_{3}+2H_{2}O$

Explanation:

$OH^{-}$ in $Mg(OH)_{2}$ reacts with $NH_{4}^{+}$ to form $NH_{3}$ and $H_{2}O$ .

Due to this acid-base reaction, $Mg(OH)_{2}$ become soluble whenever $NH_{4}NO_{3}$ is added to suspension of $Mg(OH)_{2}$ .

In this reaction, $NH_{4}NO_{3}$ acts as an acid and $Mg(OH)_{2}$ acts as a base.

Molecular equation: $Mg(OH)_{2}+2NH_{4}NO_{3}\rightarrow Mg(NO_{3})_{2}+2NH_{3}+2H_{2}O$

Total ionic: $Mg(OH)_{2}+2NH_{4}^{+}+2NO_{3}^{-}\rightarrow Mg^{2+}+2NO_{3}^{-}+2NH_{3}+2H_{2}O$

Net ionic: $Mg(OH)_{2}+2NH_{4}^{+}\rightarrow Mg^{2+}+2NH_{3}+2H_{2}O$

7 0

3 years ago

What does it mean for an acid and base to neutralize each other?

ruslelena [56]

This means that they mixed .

6 0

3 years ago

Given the reaction: solid sodium reacts with chlorine gas to form solid sodium chloride How many moles of the product result fro

lyudmila [28]

Answer:

number of moles of NaCl produce = 12 mol

Explanation:

Firstly, we need to write the chemical equation of the reaction and balance it .

Na(s) + Cl2(g) → NaCl(s)

The balanced equation is as follows:

2Na(s) + Cl2(g) → 2NaCl(s)

1 mole(71 g) of chlorine produces 2 moles(117 g) of sodium chloride

6 mole of chlorine gas will produce ? mole of sodium chloride

cross multiply

number of moles of NaCl produce = 6 × 2

number of moles of NaCl produce = 12 moles

number of moles of NaCl produce = 12 mol

3 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago