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A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

Naddika [18.5K]

Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span> $molality =n(solute) / m(solvent) =\ \textgreater \ 4.8m = 4.8mole / 1kg$
= 4.8 mole * 98 g/mole = 470g
$m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \$ $=\ \textgreater \ Molarity = 4.8 mole / 1.2 L = 4 M$ m(H2SO4) which is =<span>470g

</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
$Molarity = n(solute) / V(solution) =\ \textgreater \$

5 0

3 years ago

A bottle has a capacity of 1.2 liters. If the density of ether is 0.74 g/mL, what mass of ether can the bottle hold?

AfilCa [17]

Answer:

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance. We calculate as follows:

Volume = 60.0 g ( 1 mL / 0.70 g ) = 85.71 mL

Therefore, the correct answer is option B.

Explanation:

8 0

3 years ago

Someone help please :)

zlopas [31]

6.79 x 10⁶

<h3>Further explanation </h3>

Scientific Notation consists of 2 parts

1. the first part: the first digit followed by the decimal point

2. The second part: the power of 10 (the exponents).which shows the number of digits after the decimal point

To add/subtract, the exponents must be the same and then add/subtract the coefficient

convert the largest exponent

$\tt 7.2\times 10^6\rightarrow 72\times 10^5$

subtract the coefficients

$\tt =(72-4.1)\times 10^5\\\\=67.9\times 10^5\rightarrow 6.79\times 10^6$

7 0

3 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

How many sites on this antibody molecule have potential to bind to a non-self molecule?

Svetradugi [14.3K]

There are two sites on the antibody molecule that have a potential to bind to a non-self molecule. The Fab of an antibody is the region of the antibody that binds to an antigen. It consist of one constant and one variable domain from each heavy and light chain of the antibody. During immune reaction, an antigen-antibody reaction occurs between the antibodies made by the B cells and the antigens.

7 0

3 years ago