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3 years ago

Please help, I'll give brainiest!

Chemistry

1 answer:

Whitepunk [10]3 years ago

8 0

is it a word problem beacuse i cant see the image. Im really good a chem but if you can write the question in the commets il answer it in the comments

Explanation:

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I need this ASAP thank you

son4ous [18]

The first one I god it wrong and he told us the answer

4 0

3 years ago

A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is conve

Helen [10]

Answer:

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

Explanation:

The differential equation is given as:

$\frac{dC}{dt} = r- kC$

$\frac{dC}{r- kC} = dt$

Taking integral of both sides; we have:

$\int\limits \frac{dC}{r- kC} = \int\limits dt$

$-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD$

$r-kC=e^{-kt-kD}$

$r-kC=e^{-kt}e^{-kD}$

$r-kC=Ae^{-kt}$

$kC=r-Ae^{-kt}$

$C=\frac{r}{k}-\frac{A}{k}e^{-kt}$

$C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}$ ------- equation (1)

If C(0)= $C_o$ ; we have:

C(0)= $\frac{r}{k}-\frac{A}{k}e^0$ (where; A is an integration constant)

$C_o = \frac{r}{k}- \frac{A}{k}$

$C_o=\frac{r-A}{k}$

$kC_o=r-A$

$A=r-kC_o$

Substituting $A=r-kC_o$ into equation (1); we have;

$C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}$

3 0

4 years ago

CO_2 sublimes readily at 25°C. Which properties are usually associated with a compound that undergoes this kind of change?

antoniya [11.8K]

Explanation:

Sublimation is defined as a process in which solid state of a substance directly changes into vapor or gaseous state without undergoing liquid phase.

For example, naphthalene balls show sublimation at room temperature.

As this process does not cause any change in chemical composition of a substance. Hence, it is known as a physical process.

Similarly, when $CO_{2}$ sublimes readily at $25^{o}C$ . This shows change in physical state of carbon dioxide is taking place, i.e, from solid to gaseous phase.

Thus, we can conclude that when $CO_{2}$ sublimes readily at $25^{o}C$ then it means physical properties are usually associated with a compound that undergoes this kind of change.

4 0

3 years ago

Nail polish remover containing acetone was spilled in a room 7.2 m × 5.8 m × 3.9 m. Measurements indicated that 6,400 mg of acet

sergeinik [125]

Volume of room = $7.2 m \times 5.8 m \times 3.9 m$

= $162.864 m^{3}$

Now, according to conversion factor, convert 6,400 mg to micrograms

Since, 1 mg is equal to 1000 microgram.

Therefore, 6,400 mg = $6,400 \times 1000$

= 6,400,000 micrograms of acetone.

To calculate concentration of acetone, divide volume and evaporated amount of acetone in micrograms.

Thus,

Concentration of acetone = $\frac{6,400,000 microgram}{162.864 m^{3}}$

= 39296.5910 microgram per cubic meter or $3.92\times 10^{4} \mu g/m^{3}\simeq 4.0\times 10^{4} \mu g/m^{3}$

Hence, concentration of acetone is equal to $4.0\times 10^{4} \mu g/m^{3}$

7 0

3 years ago

A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde

hram777 [196]

The pH of the solution after addition of 28 mL of NaOH is added to HCl is $\boxed{{\text{0}}{\text{.85}}}$ .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

${\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}$

......(1)

Rearrange equation (1) to calculate the moles of solute.

${\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)$ ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}$

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

$\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}$

The reaction between HCl and NaOH occurs as follows:

${\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}$

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

$\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}$

The volume after the addition of NaOH can be calculated as follows:

$\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}$

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

$\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}$

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

${\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]$ ......(3)

Here,

$\left[{{{\text{H}}^ + }}\right]$ is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of also becomes 0.141 M.

Substitute 0.141 M for $\left[{{{\text{H}}^ + }}\right]$ in equation (3).

$\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}$

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

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3 years ago

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