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ira [324]
4 years ago
6

What is the experimental yield of Li2O?​

Chemistry
1 answer:
prohojiy [21]4 years ago
5 0
An excess of lithium oxide undergoes a synthesis reaction with water to produce lithium hydroxide​Li2O+H2O→2LiOHIf 1.05 g of water reacted
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How much heat is required to raise the<br> temperature of 835g of water from 35.0°C to<br> 65.0°C?
Anna007 [38]

Answer:

25,050 calories.

Explanation:

A calorie is the amount of energy needed to raise the temperature of one gram of water 1 degree centigrade. If we are raising 835 grams of water 30 degrees then we multiply 835*30 to get 25,050 calories.

3 0
3 years ago
Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the
Norma-Jean [14]

Answer:

9.18g

Explanation:

Step 1: Write the reduction half-reaction

Au³⁺(aq) + 3 e⁻ ⇒ Au(s)

Step 2: Calculate the mass of gold is produced when 15.0A of current are passed through a gold solution for 15.0min

We will use the following relationships:

  • 1 min = 60 s
  • 1 A = 1 C/s
  • 1 mole of electrons has a charge of 96486 C (Faraday's constant).
  • 1 mole of Au is produced when 3 moles of electrons circulate.
  • The molar mass of Au is 196.97 g/mol.

The mass of gold produced is:

15.0 min \times \frac{60s}{1 min} \times \frac{15.0C}{1s} \times \frac{1 mol e^{-} }{96486C} \times \frac{1molAu}{3 mol e^{-} } \times \frac{196.97gAu}{1molAu} = 9.18gAu

3 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
What is it called when a girl is romantically and sexually attracted to women but find men attractive?
umka21 [38]

Answer:

i think is biromantic, homosexual woman is romantically attracted to people of multiple genders, but is only sexually attracted to women. A biromantic, homosexual man is romantically attracted to people of multiple genders, but is only sexually attracted to men.

Explanation:

<em>i hope this help :/</em>

5 0
3 years ago
Read 2 more answers
A sodium bromide solution is added to a beaker containing aqueous chlorine. What would happen?
Darya [45]

Answer:

See detailed explanation.

Explanation:

Hello!

In this case, for the described chemical reaction, we can proceed as follows:

A) For the complete chemical reaction we note down every reacted and produced species as well as the proper balancing process:

2NaBr(aq)+Cl_2(aq)\rightarrow 2NaCl(aq)+Br_2(g)

In which gaseous bromine may give off.

B) The dissociated ionic equation requires the ionization of the aqueous species in ions, expect for chlorine which is not ionized:

2Na^+(aq)+2Br^-(aq)+Cl_2(aq)\rightarrow 2Na^+(aq)+2Cl^-(aq)+Br_2(g)

C) For the net ionic equation we cancel out the sodium ions as they are at both reactants and products:

2Br^-(aq)+Cl_2(aq)\rightarrow +2Cl^-(aq)+Br_2(g)

D) Based on C) we infer that the spectator ions here are the sodium ions.

Best regards!

6 0
3 years ago
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