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1 year ago

How many moles of lead (ii) chromate are in 51 grams of this substance? answer in units of mol?

Chemistry

1 answer:

maria [59]1 year ago

5 0

Mass of lead (II) chromate is 51 g. The molecular formula is $PbCrO_{4}$ and its molar mass is 323.2 g/mol

Number of moles can be calculated using the following formula:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Putting the values,

$n=\frac{(51 g}{323.1937 g/mol}=0.1578 mol$

Therefore, number of moles of lead (II) chromate will be 0.1578 mol.

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Marie mixed 5g of carbon with 5g of lead oxide. she heated the mixture strongly for 15 minutes in a fume cupboard.

olga2289 [7]

The problem talks about two questions and these are:

1. Metals are very good conductors of electricity and heat. Directing heat is easier. So let Marie heat the beads and also have heat another substance, for instance, water. If the beads heat quicker, then they are metals. Another test to conduct is called flame test. This test should give you a colored flame (blue/white for lead) the metal is lead if the reaction is: 2PbO+C ==> 2Pb +CO2

2. The beads are possibly to be lead since Ferrous(lead) oxide + carbon = carbon dioxide + lead

7 0

1 year ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

1 year ago

How many moles of oxygen atoms are in 132.2 g of MgSO4?

zzz [600]

4.4moles of oxygen atoms

Explanation:

Given parameters:

Mass of MgSO₄ = 132.2g

Unknown:

Number of moles of oxygen atoms = ?

Solution:

The number of moles is the quantity of substance that contains the avogadro's number of particles.

To solve for this;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of MgSO₄ = 24 + 32 + 4(16) = 120g/mole

Number of moles = $\frac{132.2}{120}$ = 1.1 moles

1 moles of MgSO₄ we have 4 moles of oxygen atoms

1.1 moles of MgSO₄ contains 4 x 1.1 moles = 4.4moles of oxygen atoms

learn more:

number of moles brainly.com/question/1841136

#learnwithBrainly

8 0

2 years ago

Complete the following math problem and round your answer to the correct number of significant figures. Explain why your answer

BigorU [14]

Answer:

file link http/www.openfree

7 0

1 year ago

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$

So the total mole of chloride ions:

$N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-$

And the total volume by adding the volume of each solution in L:

$V=0.500L+0.425L=0.925L$

Finally, the molarity turns out:

$M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M$

Best regards.

5 0

2 years ago