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3 years ago

Take off contact lenses first before using an eye wash station . True or False

Chemistry

2 answers:

katen-ka-za [31]3 years ago

4 0

A true because whatever you’re washing out, will get stuck behind your lenses

ryzh [129]3 years ago

3 0

Answer:

TRUE

Explanation:

You might be interested in

How many atoms of aluminum are in one formula unit of aluminum chloride?

ICE Princess25 [194]

The formula for aluminum fluoride is $AlF_{3}$

That means that there is 1 Aluminum and 3 fluorde ions in 1 molecule. The 3 does not effect the Aluminum at all.

The Answer is A

7 0

3 years ago

a student is holding a model of the human brain whht are two benefits of this model? side note: HELP IT'S A TEST AND I ONLY HAVE

mojhsa [17]

Answer:

One, It can help children to easily learn or recognize the structure of the brain.

Second, It can help children to learn about the ridges and groves, blood vessels in the brain which will grow curiosity in them to do more research about the stucture and function of the brain.

Explanation:

I hope this helps :)

6 0

3 years ago

How many molecules are in 4.92 moles of water?

bezimeni [28]

Answer:

Number molecules present = 4.92 multipled by avogadro's number.

4 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.

4 0

4 years ago

Read 2 more answers

A 312 g sample of a metal is heated to 277.845 °C and plunged into 200 g of water at a temperature of 11.945 °C. The final tempe

tigry1 [53]

Answer:

The specific heat capacity of the metal is 1.307 J/g °C

Explanation:

<u>Step 1:</u> Data given

mass of the metal = 312 grams

initial temperature of the metal ( before plunged in the water) = 277.845 °C

initial temperature water = 11.945 °C

Final temperature of water (and aslo the metal) = 99.062 °C

Specific heat capacity of wayer = 4.184 J/g °C

<u>Step 2:</u> Calculate the specifi heat capacity of the metal

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(metal) * C(metal) * ΔT(metal) = - m(water) * C(water) * ΔT(water)

with mass of metal = 312 grams

with C(metal) = TO BE DETERMINED

with ΔT (metal) = 99.062 - 277.845 = -178.783 °C

with mass of water = 200 grams

with C(water) = 4.184 J/°C * g

with ΔT (water) = 99.062 - 11.945 = 87.117 °C

312 * C(metal) * -178.783 = - 200* 4.184 * 87.117

C(metal) = -72899.51 / 55780.296 = 1.307

The specific heat capacity of the metal is 1.307 J/g °C

6 0

3 years ago