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3 years ago

Im literally failing some classes can some one help me with them

Chemistry

2 answers:

polet [3.4K]3 years ago

6 0

Answer:

What are your subjects you need help with?

galben [10]3 years ago

5 0

Answer:in can

Explanation:

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Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.

How many moles of MgO will be produced?

0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.

What's the mass of 0.541284 moles of MgO?

Formula mass of MgO:

$24.301 + 15.999 = 40.300\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of 0.541284 moles of MgO:

$m = n \cdot M = 0.541284\times 40.300 = 21.8\;\text{g}$ .

7 0

3 years ago

Which wave has the greatest distance from trough to trough?

mylen [45]

Answer:

Explanation:

8 0

3 years ago

A student sees a change in mass in the lab when mixing two chemicals what is a possible explanation?

Hitman42 [59]

Answer:

The mixing of two chemicals may result in the production of a gas which is lost to the air. This will reduce the mass of the chemical mixture, because mass is being lost in a gaseous form.

4 0

3 years ago

For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a

scoray [572]

Answer: The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}$

Given: $-\frac{d[I^-]}{dt}]$ = $2.4\times 10^{-3}mol/Ls$

$+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls$

The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

6 0

3 years ago

Using the standardized NaOH solution in the previous

geniusboy [140]

Answer:

Using the standardized NaOH solution in the previous question, the titration of 5.00 mL of vinegar required 42.25 mL of NaOH. What is the % of acetic acid in the vinegar? Concentration of NaOH from previous question= 0.0986 M. David C

Explanation:

Eh?

6 0

3 years ago

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