Can someone help me real quick?

Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe:

sladkih [1.3K]

Answer:

8.59 g

2.25 g

Explanation:

According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-

Moles of Pb(OH)CL is

$= \frac{Mass}{Molar\ mass}$

$= \frac{10.0 g}{259.65g / mol}$

= 0.0385 mol

Mass of PbO needed is

$= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}$

After solving the above equation we will get

= 8.59 g

Mass of NaCL needed is

$= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}$

After solving the above equation we will get

= 2.25 g

Therefore we have applied the above formula.

7 0

3 years ago

C₃H₈ (g) + 5O₂ (g) --> 3CO₂ (g) + 4H₂O (l) would be best classified as which type of reaction?

Dennis_Churaev [7]

Answer answer : combustion

7 0

2 years ago

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

6 0

3 years ago

What isotope name has a mass number of 201

Svetlanka [38]

Answer:

Thallium is an isotope name that has a mass number of 201

3 0

4 years ago

What do scientists use to design their experiments?

velikii [3]

I believe the scientific method

4 0

4 years ago