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Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?

9966 [12]

Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+

From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3

Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3

4 0

3 years ago

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In the reaction fecl2 + 2naoh fe(oh)2(s) + 2nacl, if 6 moles of fecl2 are added to 6 moles of naoh, how many moles of fecl2 woul

pychu [463]

The balanced chemical reaction would be

<span>fecl2 + 2naoh = fe(oh)2(s) + 2nacl

Initial amounts of the reactants are given, so, we need to determine which of the reactants is the limiting reactant and use this amount to determine what is asked. However, what is being asked is how many of the FeCl2 is used in the reaction, showing that it is NaOH that is the limiting reactants. Thus, we just use the initial amount of NaOH and relate the substances by the chemical reaction as follows:

6 mol NaOH ( 1 mol FeCl2 / 2 mol NaOH ) = 3 mol FeCl2

Therefore, 3 moles of FeCl2 is used up and 3 moles of FeCl2 is also left after the reaction.</span>

5 0

3 years ago

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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

Which term best describes the rate at which global erosion takes place

dlinn [17]

The correct answer is slow

5 0

3 years ago

If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?

tester [92]

Answer:

441.28 g Oxygen

Explanation:

The combustion of hydrogen gives water as the product.
The equation for the reaction is;

2H₂(g) + O₂(g) → 2H₂O(l)

Mass of hydrogen = 55.6 g

Number of moles of hydrogen

Moles = Mass/Molar mass

= 55.6 g ÷ 2.016 g/mol

= 27.8 moles

The mole ratio of Hydrogen to Oxygen is 2:1

Therefore;

Number of moles of oxygen = 27.5794 moles ÷ 2

= 13.790 moles

Mass of oxygen gas will therefore be;

Mass = Number of moles × Molar mass

Molar mass of oxygen gas is 32 g/mol

Mass = 13.790 moles × 32 g/mol

<h3> = 441.28 g</h3><h3>Alternatively:</h3>

Mass of hydrogen + mass of oxygen = Mass of water

Therefore;

Mass of oxygen = Mass of water - mass of hydrogen

= 497 g - 55.6 g

6 0

4 years ago

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