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anygoal [31]
3 years ago
8

The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

reviation (w/v) is also common. How many grams of sucrose are needed to make 725 mL of a 35.0% (w/v) sucrose solution?
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
3 0

Answer:

\boxed{\text{254 g}}

Explanation:

\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}

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What is the volume, in milliliters, occupied by 30.07 g of an object of density equal to
den301095 [7]

Answer:

\boxed {\tt 20.317567567568 \ mL}

Explanation:

The density formula is:

d=\frac{m}{v}

Let's rearrange the formula for v. the volume. Multiply both sides by v, then divide by d.

d*v=\frac{m}{v}*v

d*v=m

\frac{d*v}{d}=\frac{m}{d}

v=\frac{m}{d}

The volume can be found by dividing the mass by the density. The mass of the object is 30.07 grams and the density is 1.48 grams per milliliter.

m= 30.07 \ g\\d= 1.48 \ g/mL

v=\frac{30.07 \ g}{1.48 \ g/mL}

Divide. Note, when dividing, the grams, or g will cancel out.

v= \frac{30.07}{1.48 \ mL}

v=20.317567567568 \ mL

The volume of the object is 20.317567567568 milliliters.

3 0
3 years ago
The melting point of a substance is a ________ property of the substance
4vir4ik [10]
Chemical property of the substance but then again it could be physical

8 0
3 years ago
Can someone please definition of atomic radius using , nucleus, valence electrons and energy.❤️
OLEGan [10]

Answer:

Explanation:

-The Atomic Radius of an element is the distance between the center of an atom

-nucleus and its outermost, or valence electrons. ... These changes are caused by the interaction between the positive charge of the protons

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5 0
4 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
Calculate the pH of a solution that is prepared by dissolving 0.23 mol of hydrofluoric acid (HF) and 0.57 mol of hypochlorous ac
FinnZ [79.3K]

Answer:

The equilibrium concentrations of HF = 0.058 , F2 = 0.006M , HClO =0.16M , and ClO2 = 7.7 × 10⁻⁷M.

Explanation:

The Ka values for HClO₃ and HF are given as 2.9 × 10⁻⁸ and 6.6 × 10⁻⁴ respectively. The molar concentration for HF = 0.23/ 3.60L = 0.064 M and 0.57/ 3.60 = 0.16 M.

When HF is reacted with water, it ionizes to form H₃O⁺ and F⁻. The concentration of H₃O⁺ and F⁻ can be calculated below:

HF(aq) <------------------------> H30^+ + F^-.

Ka = [H^+] [F^-]/[HF] .

6.6× 10^-4 = [x][x]/ ( 0.064- x).

x = 0.0060 M.

The concentration of H₃O⁺ and F⁻ = 0.0060 M respectively.

The pH = - log [ H₃O⁺ ] = -log [0.0060] = 2.22.

When HClO is reacted with water, it ionizes to form H₃O⁺ and F⁻. The concentration of H₃O⁺ and ClO⁻ can be calculated below:

HClO(aq) <------------------------> H30^+ + ClO^-.

Ka = [H^+] [ClO^-]/[HClO] .

6.6× 10^-4 = [0.006 + x] [x]/ ( 0.16 - x).

x = 7.7 × 10^-7M.

[ClO^-] = 7.7 × 10^-7 M.

[HClO] = 0.16 - 7.7 × 10^-7 = 0.16M.

[F^-] = 0.006 M.

[HF] = 0.064 - 0.006 = 0.058 M.

5 0
3 years ago
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