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3 years ago

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Magnesium and Hydrochloric acid (HCl) react to form magnesium chloride (MgCl2) and hydrogen gas. If you started with 65.8 g of m

agnesium and 45.7 g of hydrochloric acid. What is the limiting reactant? Ensure you post calculations to prove your answer. (3 marks)

1 answer:

DiKsa [7]3 years ago

4 0

Answer:

Cl

Explanation:

n for Mg=m:Mm

=65.8:24

=2.74mol

n for Cl=m:Mm

=45.7:70

=0.65mol

According to the <em>number of moles</em> we can say that the limiting reagent is Cl as it has few moles.

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Hunting lions has been banned in some parts of Africa. If this law is enforced, what can we expect to happen to the populations

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The population would decrease, the more predators there are the more food needed for the species .

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3 years ago

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How many grams of oxygen are produced when 1.05 moles of hydrogen gas is<br> produced?

vivado [14]

Answer:

1.058337 grams of hydrogen and 2H2 + O2 ==> 2H2O hydrogen peroxide

mols H2 = grams/molar mass

Using the coefficients in the balanced equation, convert mols H2 to mols O2.

Now convert mols O2 to grams. That's grams mols O2 x molar mass O2.

and it could also produce H2O water but no air but it could make other things

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3 years ago

The ph of a solution for which [OH]= 1.0 x 10^-4 is?<br><br> A 10<br> B 14<br> C 11<br> D 9<br> E 7

zhenek [66]

To find pH, use the following formula ---> pH= - log [H+]

so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in

[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10

now that we have the H+ concentration, we can solve of pH

pH= -log (1.0x10-10)= 10

answer is A

8 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

What is the pH of a solution in which [H 3O +] = 3.8 × 10 -8 M?

7nadin3 [17]

Answer:

<h3>The answer is 7.42 </h3>

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ { H_3O}^{+}]$

From the question we have

$pH = - log(3.8 \times {10}^{ - 8} ) \\ = 7.420216...$

We have the final answer as

<h3>7.42 </h3>

Hope this helps you

4 0

3 years ago