Discuss the molar equivalents of sodium borohydride. does this make sodium borohydride the limiting reagent? why or why not. (a couple sentences are sufficient).

Answer:

okay

Explanation:

The fire triangle's three sides illustrate the three elements of fire: heat, fuel and oxidization. The three elements must be combined in the right proportions for a fire to occur. If any of the three elements are removed, the fire is extinguished

(a) barium and oxygen - Braium oxide ([tex]BaO[/tex])

(b) calcium and phosphorus - Calcium phosphate - ([tex]Ca_{3}(PO_{4})_{2}[/tex])

(c) aluminum and sulfur - Aluminum Sulfide [tex]Al_{2}S_{3}[/tex]

lithium and nitrogen - Lithium nitride [tex]Li_{3} N[/tex]

The term ionic compound refers to a compound that is formed when ions are combined together. We know that ionic compounds are crystalline and that they are also soluble in water and polar solvents.  Most importantly, the ions are arranged in a very definite pattern.

Now let us see the names of the compounds formed by each pair;

(a) barium and oxygen - Braium oxide ([tex]BaO[/tex])

(b) calcium and phosphorus - Calcium phosphate - ([tex]Ca_{3}(PO_{4})_{2}[/tex])

(c) aluminum and sulfur - Aluminum Sulfide [tex]Al_{2}S_{3}[/tex]

lithium and nitrogen - Lithium nitride [tex]Li_{3} N[/tex]

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The empirical formula of the magnesium oxide formed is MgO

Here given data is piece of magnesium with a mass = 5.867 g  after the magnesium had combusted and allowed to cool down the product and mass = 8.956 g

The empirical formula is the simplest whole number ratio of atoms in a compound and the ratio of atoms is the same as the ratio of moles and when crucible react with magnesium and air means oxygen then magnesium reacts readily with oxygen in the air, to produce magnesium oxide then chemical reaction is written as follows:

2Mg + O₂ → 2MgO

and hence when it cool down automatically the mass of magnesium is more than 1st mass because in that oxygen of air are added and that's why MgO are made the empirical formula of the magnesium oxide formed is MgO

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6.508 g mass of of NH3 is formed when 5.36 g of nitrogen gas reacts with excess hydrogen gas.

                                N2 + 3 H2  → 2 NH3

The molar mass of N2 is 28.0 g/mol. The molar mass of NH3 is 17 g/mol.

To find the mass of NH3:

=5.36 x 2 x (17 / 28)

=182.2 / 28

= 6.508 g

The mass of ammonia is 5.36 g.

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6 CO2 molecules are involved in this reaction which is in turn helping to produce 1 molecule of glucose.

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Answer:

Water dissolves with water because both are polar solvents

[tex].[/tex]

The Density of the given object is 1.61gm/cm³.

Given,

mass of the object = 1072g

Length of the object = 15.2 cm

width of the object = 6 cm

height of the object = 7.3 cm

Therefore volume of the object V = l × w × h

V = 15.2 × 6 × 7.3

= 665.76 cm³

As we know,

Density = Mass / Volume

Density = 1072 / 665.76

= 1.61 g/cm³

Therefore, the density of given object is 1.61 g/cm³.

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The name given to an ion that has lost electrons is cation. Details about cations can be found below.

A cation is a positively charged ion i.e. the one that would be attracted to the cathode in an electrolytic procedure.

Generally, an ion is an electrically charged atom. An ion can either be positively charged or negatively charged. The latter is called an anion while the former is called a cation.

A cation is formed when an atom loses electrons to become positively charged while an anion is formed when an atom accepts electrons to become negatively charged.

Therefore, it can be said that the name given to an ion that has lost electrons is cation.

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The reagents need for the formation of 2-ethyl heptanoic acid are Bromoethane, 1-bromopentane, sodium ethoxide, and potassium tert-butoxide.

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The molality of the glycerol solution is 3.854mol/kg

The number of moles of solute present in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.

Given:

Mass, m1 of glycerol = 15.01g

Mass, m2 of ethanol = 42.3g

Molar mass, mm1 of glycerol = 92.095g/mol

Molar mass, mm2 of ethanol = 46.069g/mol

To find:

Molality, m = ?

Formula:

Molality, m = moles of solute / kg of solvent

Calculations:

No. of moles of glycerol = 15.01 x (1/92.095)

n = 0.163 mols

Molality, m = 0.163 / 0.0423g

Molality, m = 3.854mol/kg

Result:

3.854mol/kg is the molality of the glycerol solution.

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There ought to be 6.3 grams of oxygen reacting with the fuel in accordance with the law of conservation of matter.

Fuel mass equals 20.8 grams

In the presence of oxygen, 9.8 grams of CO2 and 17.3 grams of water are generated.

The mass of the products and reactants of a chemical reaction must be equal, in accordance with the law of conservation of mass.

In other words, mass fuel plus mass oxygen equals mass carbon dioxide plus mass water.

9.8 grams of carbon dioxide per mass.

17.3 grams of water in mass

27.1 grams of mass CO2 plus water.

The fuel's mass is 20.8 grams.

This indicates that the mass of the oxygen should be 27.1 grams in addition to the mass of the fuel.

Mass oxygen is equal to 27.1 - 20.8 grams= 6.3 grams.

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Sodium is an example of an alkali metal that reacts with water to form hydrogen gas and an alkaline solution.

Alkaline metals are lithium, sodium (Na), rubidium, potassium, cesium. They have low ionization energy and lose one electron very easy.

Atomic number of sodium is 11, it means that it has 11 protons and 11 electrons, so atom of sodium is neutral.

Electron configuration of the sodium atom: ₁₁Na 1s² 2s² 2p⁶ 3s¹.

Balanced chemical equation for the reaction of sodium with water:

2Na + 2H₂O → 2NaOH + H₂

NaOH is strong alkaline base.

H₂ is the hydrogen gas

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Answer:

B) 27.94

Explanation:

All other last digits are wrong except b

Answer:

[tex] \frac{39.25 + 150.}{6.773} = 27.94182[/tex]

Explanation:

[tex]1. \: (39.25 + 150.) \div (6.60 + 0.173) \\ 2. \: (39.25 + 150.) \div 6.773 \\ 3. \: (39.25 + 150.) \times \frac{1}{6.773} \\ 4. \: \frac{(39.25 + 150.) \times 1}{6.773} \\ 5. \: \frac{39.25 + 150.}{6.773} = 27.94182[/tex]

Answer:

0.005 moles

Explanation:

120cm^3 = 0.12dm^3

1mol is always 24dm^3 of gas at rtp

0.12 ÷ 24 = 0.005 moles

Different substances are given as SO₂, HNO₃, BaCO₃, CO₂, Zn(NO₃)₂, H₃PO₄

KOH reacts with SO₂ at a lower temperature,

KOH + SO₂ → K₂ SO₃ + H₂ O

Nitric acid (an acid) and potassium hydroxide (a base) combine to form potassium nitrate (salt) and water.

KOH + HNO₃ → KNO₃ + H₂O

When substance potassium hydroxide and Barium carbonate,

KOH + BaCO₃ → K₂ CO₃ + Ba(OH)₂

Carbon dioxide and potassium hydroxide react to produce potassium carbonate and water. Carbon dioxide is taken out of the space station using potassium hydroxide.

KOH + CO₂ → K₂ CO₃ + H₂O

Potassium hydroxide reacts with zinc nitrate,

Zn(NO₃)₂ + KOH → Zn(OH)₂ + KNO₃

Potassium phosphate and water are created when potassium hydroxide and phosphoric acid combine.

KOH (aq) + H₃ PO₄ (aq) → K₃ PO₄ (aq) + H₂O (l)

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Answer:

C. To work on the mechanics of a pitchers throw
or... A. To design a stadium with the maximum number of seats

Explanation:

Answer:

All particles of matter are always in constant motion. In this case, the particles of the rock possess kinetic energy as they vibrate in place. However, the particles also contain potential energy due to their position and arrangement. This form of stored energy is responsible for keeping the particles bonded together.

Explanation:

The melting and boiling and density is given below :

Substance          melting point / °C         boiling point / °C       density / g cm⁻³

  E                           -259                                 -253                          0.09

 F                           1085                                 2580                         8.93

  G                                 -7                                     59                            3.1

 H                              -39                                   357                           13.6

  I                                -218                                  - 183                       0.0013

A) substances are gases at room temperature:

    E and I are gases at room temperature because the melting and boiling point are below at room temperature.

B) substances are liquids at room temperature:

    G and H  are liquids at room temperature because substances are liquid at room temperature if melting point is below room temperature and boiling point is above room temperature.

C) substances are solid at room temperature :

   F is solid at room temperature because its melting point is above room temperature.

D) substances are most likely to be metal :

   F is metal because metals have very high melting and boiling point except mercury.

E) substance are most likely to be mercury :

    H is most likely to be mercury because mercury has boiling point -38°C and melting point is 356°C and density is 13.5 g/ cm³ and mercury is metal with exception.

G) substances will be liquid at -210°C :

  I is liquid at -210°C because substance is liquid when its melting point is below temperature and boiling point is above temperature.

H) substances are most likely to be bromine :

  G is more likely to be bromine because melting and boiling point and density of G is close to bromine.

I)  substances which are least dense non metal :

   E is least dense non metal because its melting and boiling point and density is close to least dense non - metal.

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The empirical formula of the compound is C₃H₆O

The empirical formula is determined as follows from the equation of the reaction as follows:

mass of CO₂ produced = 15.17 mg = 0.01517 g

moles of C in the sample = 0.01517/44  = 0.000344 moles

mass of C in the sample =  0.0000344  * 12 = 0.00413 g

mass of H₂O produced  = 6.19 mg = 0.00619 g

moles of H in the sample = 0.00619/18 * 2 = 0.000688 moles

mass of H in sample = 0.000688 g

mass of oxygen in the sample = 0.00667 g - (0.00413 g + 0.000688 g)

mass of oxygen in the sample = 0.00185

moles of oxygen = 0.00185/16 = 0.000116

Ratio of C : H : O = 0.000344 : 0.000688 : 0.000116

Ratio of C : H : O = 3 : 6 : 1

Empirical formula of compound = C₃H₆O

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A high amount of naturally occurring NaCl is considered evidence of a compound with a low activation energy and is denoted as option A.

This is referred to as the minimum amount of energy which is needed for a chemical reaction to occur and is dependent on various factors.

A compound which has a low activation energy will have a high yield sue to the ease in which the reactants are being converted into products. in this scenario, the high amount of naturally occurring NaCl signifies a low activation energy which is why option A was chosen as the correct choice.

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Yes, to determine the molar mass of an unknown solute from the freezing point depression the freezing points of both the pure solvent and the solution need to be measured.

The depression of the freezing point is a and it is given by the following formula,

ΔT f = T°f - T f

Here T°f is the freezing point of pure solvent

T f is the freezing point of the solution made by adding non-volatile solute to solvent

The freezing point depression is directly proportional to the molality of the solution

T(f) ∝ m

T(f)= K f x m

Here K f is the proportionality constant.

Hence, to determine the molar mass of an unknown solute from the freezing point depression the freezing points of both the pure solvent and the solution need to be measured.

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41296.22 g of weight of methanol (ch3oh) must be added to water to prepare 191.8 ml of a solution that is 6.72 m methanol

It is given that 191.8 ml of 6.72 M  methanol is to be prepared.

Using the Molarity formula , we get

                 6.72 = n / 191.8

                    n = 6.72 × 191.8

                    n = 1288.89 g / mol

Molar mass of methanol = 32.04g/mol

Weight of methanol to be added= 32.04 ×1288.89 =  41296.22 g

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Answer:

Because it is the same element but has different atomic mass

For the reaction 4ch3ch3 14no2, the rate of the reaction in terms of the change in nitrogen concentration would be written as:+ (2/3) Δ[H2O]/Δt.

The chemical element with the atomic number 7 and the letter N is known as nitrogen. The lightest element in group 15 of the periodic table, often known as the pnictogens, nitrogen is a nonmetal. It is a typical element in the cosmos, believed to be eighth in the Milky Way and the Solar System in terms of overall abundance. Two of the element's atoms combine to generate N2, an odorless and colorless diatomic gas, under standard pressure and temperature.

The majority of the atmosphere on Earth is made up of N2, making it the most prevalent uncombined element. All living things contain nitrogen, which is principally found in the amino acids that make up proteins, the nucleic acids that make up DNA and RNA, and the energy-transfer molecule adenosine triphosphate.

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When H₂SO₄ is neutralized by NaOH in an aqueous solution, the net ionic equation is:

2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l)

A chemical equation known as an ionic equation expresses the electrolytes in an aqueous solution as dissociated ions. For the reaction between [tex]H_{2}SO_{4}[/tex] , NaOH that neutralizes it, we must write the net ionic equation.

The reaction's equilibrium chemical equation is:

[tex]Na_{2} SO_{4}[/tex](aq) + [tex]2H_{2}O[/tex] = [tex]H_{2}SO_{4}[/tex] (aq) + [tex]2NaOH[/tex](aq) (l)

We shall now draft the ionic equation.

The reaction's ionic equation is

[tex]2 Na ^{+} (aq) + SO_{4} ^{2-} (aq) + 2H_{2}O = 2H^{+} (aq) + SO_{4} ^{2-} (aq) + 2 Na ^{+} (aq)+ 2OH(aq) (l)[/tex]

The net ionic equation is then obtained by canceling out the spectator ions Na+ and SO42.

The neutralization reaction's net ionic equation is:

2H⁺(aq),  2OH⁻(aq), and 2H₂O (l)

In light of this, the net ionic equation for the neutralization of H2SO4 and NaOH is:

2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l)

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Answer: S

:) hope this helps you!

Ask a question the first things that scientists do when starting a new investigation.

Scientific practices, which include things like answering questions, creating and using models, and having arguments, are cognitive, discursive, and social activities carried out in science classrooms with the goal of developing epistemic understanding and appreciation of the nature of science.

Participating in these activities not only makes students' knowledge more meaningful and deeply ingrained in their worldview, but it also aids them in developing an understanding of the disciplinary and transdisciplinary ideas of science and engineering.

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