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2 years ago

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Magnesium hydroxide reacts with hydrogen chloride, a double-displacement reaction occurs and produces magnesium chloride and wat

er. If you start with 525.0mg of Magnesium hydroxide, how many atoms do you have?

1 answer:

Natalija [7]2 years ago

5 0

Explanation:

magnesium Hydroxide + Hydrochloric react together and give us magnesium chloride + water

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svlad2 [7]

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➷ It would be a covalent bond most likely. It cannot be an ionic bond as the two elements have the same number of electrons. It also couldn't be a metallic bond as they are obviously not metals.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0

3 years ago

Read 2 more answers

A solution is made by mixing equal masses of methanol, CH4O, and ethanol, C2H6O. Determine the mole fraction of each component t

Serggg [28]

Answer: mole fraction of methanol = 0.590

mole fraction of ethanol = 0.410

Explanation:

We are given:

Equal masses of methanol $CH_4O$ and ethanol $C_2H_6O$ are mixed.

let the mass be x g.

Calculating the moles of methanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{32.04g/mol}$

Calculating the moles of ethanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}$

To calculate the mole fraction of methanol, we use the equation:

$\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590$

To calculate the mole fraction of ethanol, we use the equation:

$\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410$

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

5 0

3 years ago

Consider the reaction of zinc metal with hydrochloric acid.

bearhunter [10]

2Zn(g)+2Hcl(aq)---->2ZnCl(s) +H2(g)

7 0

3 years ago

Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating

kvv77 [185]

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

$1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol$

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

$\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }$

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = $\frac{1}{1}\times 0.0295=0.0295moles$ of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g$

To calculate the percentage yield of iodo-salicylamide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

$\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%$

Hence, the percent yield of the reaction is 68.68%.

4 0

4 years ago

A 4.000 g sample of iron was heated from 0.0°C to 21.0°C. If it has a specific heat of 0.440 kJ/kg °C, how many kilojoules of he

Mashutka [201]

M = 4 g = 4/1000 kg = 0.004 kg, θ₂ = 21.0°C, θ₁ = 0°C , c = 0.44kJ/kg°C,

Q = mc(θ₂ - θ₁)

Q = 0.004*0.440*(21 - 0)

Q = 0.03696 kJ

0.03696 kJ<span> of heat is absorbed.</span>

6 0

3 years ago