Answer:
400 cm³ of ammonia, NH₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.
Finally, we shall determine the maximum volume of ammonia, NH₃ produced from the reaction. This can be obtained as illustrated below:
From the balanced equation above,
3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.
Therefore, 600 cm³ of H₂ will react to produce = (600 × 2)/3 = 400 cm³ of NH₃.
Thus, 400 cm³ of ammonia, NH₃ were obtained from the reaction.
How many moles of Cd and of N are contained in 132.4 g of Cd(N03)2-4H20? (b) How many molecules of water of hydration are in this same amount?
Answer: The given amount of [tex]Cd(NO_3)_2.4H_2O[/tex] contains 0.430 moles of Cd, 0.860 moles of N and [tex]2.59\times 10^{23}[/tex] molecules of water.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of [tex]Cd(NO_3)_2.4H_2O[/tex] = 132.4 g
Molar mass of [tex]Cd(NO_3)_2.4H_2O[/tex] = 308.5 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }Cd(NO_3)_2.4H_2O=\frac{132.4g}{308.5g/mol}=0.430 mol[/tex]
1 mole of [tex]Cd(NO_3)_2.4H_2O[/tex] contains 1 mole of Cd, 2 moles of nitrogen atom (N), 10 moles of oxygen atom (O) and 8 moles of hydrogen atom (H).
So, 0.430 moles of [tex]Cd(NO_3)_2.4H_2O[/tex] will contain = [tex](1\times 0.430) = 0.430mol[/tex] of Cd and [tex](2\times 0.430)=0.86mol[/tex] of N
According to the mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, 0.430 moles of [tex]Cd(NO_3)_2.4H_2O[/tex] will contain = [tex]\frac{6.022\times 10^{23}}{1mol}\times 0.430mol=2.59\times 10^{23}[/tex] number of water molecules.
Hence, the given amount of [tex]Cd(NO_3)_2.4H_2O[/tex] contains 0.430 moles of Cd, 0.860 moles of N and [tex]2.59\times 10^{23}[/tex] molecules of water.
9. During an experiment the students prepared three mixtures A)Starch in water B) Sodium chloride solution C) Tincture of Iodine. i) Students observed a visible beam of light through mixture A. Why? ii) Tincture of lodTe did not show Tyndall effect. Explain reason. ill) How can you relate particle size to Tyndall effect?
Answer:
See explanation
Explanation:
Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.
In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.
Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.
Describe the formation of oxygen molecule
Answer:
oxygen molecule has two oxygen atom . Each O atom share 2 electrons to form two covalent bonds out of which one is sigma bond and other is pi bond . sigma bond is formed by axial overlap 2p atomic orbitals of oxygen and pi bond is formed of lateral overlap of 2p atomic orbitals of oxygen .
What is the answer to 9.7300x10^2+9.8700x 10^3 1.0843x 10^ in scientific notation
Explanation:
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two easy uses of mixture
Explanation:
it helps to make juices.
It helps to make concentrated acid into dilute acid.
Answer:
Explanation:
Here are a few more examples:
1. Smoke and fog (Smog)
2. Dirt and water (Mud)
list three factors that affect the rate of a chemical reaction
Answer: 1) temperature, 2) add a catalyst, and 3) concentrations
Explanation: A rise in temperature adds kinetic energy to the reactants, leading to more frequent, and energetic collisions, which will often accelerate the reaqction. A catalyst is an agent that helps the reaction move forward more quickly by offering a substrate that aids in orienting the reactants for more efficient reaction. A higher concentration increases the chances of collisons that result in products.
what is the name of this organic molecule
Answer:
C2H4 Ethylene
Explanation:
Hope it helps! :)
which chloride is a coloured solid rtp
Answer:
sodium chloride
hope that helped :)
Answer:
Chlorine is a greenish yellow gas at room temperature and atmospheric pressure. ... yielding chlorine water, and from this solution a solid hydrate of ideal ...
Which of the following are held together by nonpolar covalent bonds?
The Periodic Table
A. Atoms of phosphorus (P) and chlorine (CI)
B. Atoms of chlorine (CI)
C. Atoms of chlorine (CI) and magnesium (Mg)
D. Atoms of magnesium (Mg)
Answer:
Answer: Atoms of chlorine (CI) and magnesium (Mg)
Answer: Atoms of chlorine (CI) and magnesium (Mg)is your answer
How many moles of oxygen are required to react completely with 5 mol C8H18?
Answer:
62.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.
Thus, 62.5 moles of O₂ is needed for the reaction.
what is second explain it briefly
Explanation:
it is that when an object is moving 25 m per sec to calculate its time we can simply divide 1.5 by 25 and multiply by 1
A compound was found to contain 90.6% lead (Pb) and 9.4% oxygen. What is the empirical formula for this compound?
Answer:
the answer is 47.9 and ik because I just had that question
The empirical formula of the compound is O₄Pb₃.
What is the empirical formula?
An Empirical system is the chemical system of a compound that offers the proportions of the elements gifted within the compound however not the real numbers or arrangement of atoms. This would be the lowest complete variety ratio of the elements within the compound.
Amount of lead (Pb) = 90.6%
⇒and amount of oxygen = 9.4%
taking the whole number ratio
o = 4
Pb = 3
∴ ⇒O: Pb=4:3
O4Pb3 answer.
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What will happen if we drink liquid nitrogen?
Oxide of nitrogen that is acidic
Answer:
Nitrogen oxides are used in the production of nitric acid, lacquers, dyes, and other chemicals. Nitrogen oxides are used in rocket fuels, in the nitrification of organic chemicals, and in the manufacture of explosives.
Explanation:
a. For a chemistry lab final exam, a high school chemistry student was given a 1-mole sample of CaCl2 and a 1-mole sample of MgCl2 but was not told which sample was which. He was to identify the powders.
He looked up the enthalpies of formation for both of the chemicals and calculated the ΔHreaction for dissolving each powder: CaCl2 (s) Ca 2+ (aq) + 2Cl – (aq), and MgCl2 (s) Mg 2 (aq) + 2Cl – (aq). He then put each powder in a coffee-cup calorimeter and added water.
When sample A dissolved, the temperature increased by 0.74°C. When sample B dissolved, the temperature increased by 0.39°C. Which chemical was A, and which was B? Use the table of enthalpies of formation to help you. Explain your reasoning.
Ca^2+ ion is smaller than Mg^2+ ion hence it has a greater heat of formation or lattice energy.
When a crystal lattice is formed, the energy that is released if the component ions of the compound are brought together from infinity is called the lattice energy. It is the energy released when a crystal lattice is formed from its component ions.
The question lets us know that the heat released by the compounds depends on the energy released upon formation of the compound. Hence, the higher the energy released upon formation, the higher the magnitude of heat released upon dissolution of the compound.
Recall that lattice energy depends on the size of the ions. Thus, the smaller the ions, the higher the lattice energy.
Ca^2+ is smaller than Mg^2+ hence more energy is given off when CaCl2 is formed than when MgCl2 is formed.
As stated above, the greater the lattice energy, the greater the heat released when the lattice dissolves and the higher the rise in temperature.
Putting all these together, Sample A must be CaCl2 while sample B must be MgCl2.
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5+64
this is TOTALLY SOOOO hard so help me
^thats what she said
Answer:
69
Explanation:
I thought that I answer is correct
Answer:
The answer is 69
Explanation:
Because if you add one its 66, and so on, hope this helps! but if you add five it will be 69, if you need help with easy equations let me know
PLZ HELP ME A girl throws a ball and her dog chases it. What characteristic of living things is this dog demonstrating as it chases the ball?
A. Cells
B. Reproduction
C. Growth
D. Response
what is the reactant(s) in the chemical equation below
3CO(g) + Fe2O3(s) 2Fe(s) + 3CO2(g)
Answer:
the reactants are carbon dioxide ,iron, oxygen
A 1.732 g sample of iron is heated in air. It oxidizes to form a product with a mass of 2.205 g.
What is the empirical formula of the product?
Answer:
A 1.732 g sample of iron is heated in air. It oxidizes to form a product with a mass of 2.205 g. What is the empirical formula of the product? FeO 4. Caffeine is made of 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen by mass.
Explanation:
The empirical formula of the product formed by the reaction of iron and oxygen is FeO.
Let's consider the following generic chemical equation:
Fe + O₂ ⇄ FexOy
According to the law of conservation of mass, the sum of the masses of the reactants is equal to the sum of the masses of the products. The mass of O₂ that reacted, and remains in the oxide is:
[tex]mFe + mO_2 = mFexOy\\mO_2 = mFexOy - mFe = 2.205 g - 1.732 g = 0.473 g[/tex]
The oxide has 1.732 g of Fe and 0.473 g of oxygen and a total mass of 2.205 g. To determine the empirical formula, we need to calculate the percent composition.
[tex]\%Fe = \frac{mFe}{mFexOy} \times 100 \% = \frac{1.732 g}{2.205 g} \times 100 \% = 78.55\%[/tex]
[tex]\%O = \frac{mO}{mFexOy} \times 100 \% = \frac{0.473 g}{2.205 g} \times 100 \% = 21.45\%[/tex]
Now, we will divide each percentage by the atomic mass of the element.
[tex]Fe: 78.55/55.85 = 1.406\\O: 21.45/16.00 = 1.340[/tex]
Finally, we divide both numbers by the smallest one.
[tex]Fe: 1.406/1.340 \approx 1 \\O: 1.340/1.340 = 1[/tex]
The empirical formula of the product is FeO.
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giúp tớ với, mình học lớp 9 ạ
Answer: broooo what is this
Explanation:
what do y’all do when y’all brainly says “locked” and won’t show anymore answers !???
Answer:
Create a new account as the thing you're experiencing now is due to the fact that Brainly had blocked your account.
Write the empirical formula for at least four ionic compounds that could be formed from the following ions: NO3, Pb^4+, NH4, SO4
Explanation:
here's the answer to your question
double arrow just mean that it's a reversible process, and the reaction can go back and forth.
Predict the approximate Ksp of Cuz(AsO4)2 based on the measured potential of Cell 7. Use the equation given in the Background.
a. 1 x 10^-35
b. 4 x 10^14
c. 5 x 10^-17
d. 2 x 10^-21
Answer:
a. 1 x 10^-35
Explanation:
The correct compound given is: [tex]\mathsf{Cu_3(AsO_4)}_2[/tex]
To predict the approximate Ksp value of the given compound, we will need to express the oxidation-reduction half-reaction of the compound and its dissociation, then, we will use the Nernst equation to determine the approximate Ksp value.
To start with the reduction half-reaction:
[tex]\mathsf{Cu_3(AsO_4)_{2}(s) + 6e^- \to 2As O_{4}^{3-}_{(aq)}+3Cu(s) }[/tex]
The oxidation half-reaction is:
[tex]\mathsf{3Cu(s) \to 3CU^{2+}_{(aq)} + 6e^-}[/tex]
The overall cell reaction now is:
[tex]\mathsf{Cu_3(AsO_4)_{2}(s) \to 3Cu^+ (aq) + 2As O_{4}^{3-}_{(aq)} }[/tex]
From the reduction half-reduction, the number of moles of electrons (n) transferred is 6 moles.
By applying the Nernst equation:
[tex]\mathsf{E_{cell} = E^0_{cell} -\dfrac{0.0591V}{n}log [Cu^{2+}]^3[AsO_4^{3-}]^2 }[/tex]
At standard conditions;
The standard cell potential [tex]\mathsf{E^0_{cell} = -0.342 \ V}[/tex]
and [tex]\mathsf{E_{cell} = 0 \ V}[/tex] since it is at equilibrium.
∴
[tex]\mathsf{0 = -0.342 -\dfrac{0.0591V}{6}log [Cu^{2+}]^3[AsO_4^{3-}]^2 } \\ \\ \\ \mathsf{0.342 = -\dfrac{0.0591V}{6}log [Cu^{2+}]^3[AsO_4^{3-}]^2 }[/tex]
[tex]\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 = \dfrac{-(0.342)*6}{0.0591 }}[/tex]
[tex]\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 = -34.7}[/tex]
[tex]\mathsf{log [Cu^{2+}]^3[AsO_4^{3-}]^2 \simeq -35}[/tex]
[tex]\mathsf{[Cu^{2+}]^3[AsO_4^{3-}]^2 = 10^{-35}}[/tex]
[tex]\mathbf{K_{sp} = [Cu^{2+}]^3[AsO_4^{3-}]^2 = 1\times 10^{-35}}[/tex]
Question 14 (Essay Worth 10 points) (03.05 MC) Use complete sentences to differentiate between acids and bases on the basis of touch. Give an example of each type.
Answer:
Acid is a molecule capable of donating hydrogen ion and they form aqueous solutions with a sour taste while base is a substance that accepts proton from proton donor and in aqueous solution, they have an astringent or bitter taste. Moreover, a good example for base is sodium hydrogen carbonate as baking soda or baking powder and for acid, the most common example is the acetic acid or vinegar.
Answer:
Well, for starters, bases mostly consist of soaps and cleaning products, acids consist of vinegar,lemon,and stuff like that. Most acids on the basis of touch can burn your skin or be very sour, unlike bases which are more soapy and have a cleaner touch.A good example of an acid is vinegar, which can be used for cleaning but can also be used for making foods, on the other hand. A good example of a base is dish soap. It is used for cleaning.
Explanation:
i just did the exam and got a 100
What is the AHsol for LiF → Lit + F-? The lattice energy is -1,036 kJ/
mol, the enthalpy of hydration for Lit is -499
kJ/mol, and the enthalpy of hydration for F-is-431 kJ/mol. Use A Hooi = -A Hat + AHhydr.
0-968 kJ/mol
-106 kJ/mol
106 kJ/mol
1,966 kJ/
mol
The heat of hydration is defined as the heat absorbed or evolved when one mole of a substance undergoes hydration. The heat of solution is the enthalpy change associated with he dissolution of a solute. The lattice energy is the heat. Lattice energy is the energy released when the components of the lattice are brought together from infinity.
Hence the heat of solution for LiF → Li^+ + F- is -1966 kJ/mol
Given that;
Heat of hydration = ΔH solution – ΔH lattice energy
Where,
ΔH solution = Heat of the solution
ΔH lattice energy = Lattice energy of the solution
The heat of solution or enthalpy of dissolution is defined as the enthalpy change associated with the dissolution of a solute in a solvent
From the formula above;
ΔH solution = Heat of hydration + ΔH lattice energy
Heat of hydration = [(-431) + (-499)] = -930 kJ/mol
ΔH solution = (-930) + (-1,036) = -1966 kJ/mol
Hence the heat of solution for LiF → Li^+ + F- is -1966 kJ/mol
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Define force and speed
Force = The external energy that changes or tends to change the state of any body or object is called force.
Speed = The rate of distance is called speed.
Answer:
Force is a push or pull which changes or tend to change the position of a body.
The rate of change its position with time or magnitude is called speed.
16. The valency of sodium is +1 and that of chlorine is -1, why?
Answer:
It because ,sodium is a metal and chlorine is a non metal
What are 5 uses of nitrogen?
Answer:
nitrogen is used in the production of 1) fertilisers 2) nitric acid 3) nylon 4) dyes and 5) explosives
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A precipitation reaction 2. A Lewis acid-base reaction the produces a coordination complex: A Lewis acid-base reaction the produces a coordination complex 3. An oxidation-reduction reaction that is also a synthesis reaction: An oxidation-reduction reaction that is also a synthesis reaction Column B a. Pb2 (aq0 CrO42-(aq) --> PbCrO4(s) b. 2 Mg(s) O2(g) -> 2 MgO(s) c. Ag (aq) 2NH3(aq) --> [Ag(NH3)2] (aq)
Answer:
Pb2+ (aq) + CrO42-(aq) --> PbCrO4(s) - A precipitation reaction
2 Mg(s) + O2(g) -> 2 MgO(s) - An oxidation-reduction reaction that is also a synthesis reaction
Ag^+ (aq) + 2NH3(aq) --> [Ag(NH3)2]^+ (aq) - A Lewis acid-base reaction the produces a coordination complex
Explanation:
A precipitation reaction is one in which two aqueous reactants yields an insoluble product called a precipitate as shown in reaction 1 above.
In the second reaction Mg is oxidized from zero to + 2 while oxygen was reduced from zero to -2 as the MgO is formed hence the reaction is an oxidation-reduction reaction that is also a synthesis reaction.
In the third reaction, Ag^+ a Lewis acid reacts with NH3, a lewis base to yield the complex [Ag(NH3)2]^+.
Sarah measures out 151 grams of SO2. How many moles is this? Express your answer to three significant figures.
Answer:
[tex]\boxed {\boxed {\sf 2.36 \ mol \ SO_2}}[/tex]
Explanation:
We are asked to convert grams to moles. We will use the molar mass and dimensional analysis to perform this conversion.
1. Molar MassThe molar mass is the mass of 1 mole of a substance. These values are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead.
We are given a mass of sulfur dioxide (SO₂). Look up the molar masses of the individual elements.
Sulfur (S): 32.07 g/mol Oxygen (O): 15.999 g/molNotice that the formula of the compound contains a subscript. The subscript after O means there are 2 moles of oxygen in 1 mole of sulfur dioxide. We must multiply oxygen's molar mass before adding sulfur's.
O₂: 15.999 * 2 = 31.998 g/mol SO₂= 32.07 + 31.998 = 64.068 g/mol2. Convert Grams to Moles
Now we will use dimensional analysis to convert grams to moles. From the molar mass, we know there are 64.068 grams of sulfur dioxide per mole, so we can set up a ratio.
[tex]\frac {64.068 \ g \ SO_2} {1 \ mol \ SO_2}[/tex]
We are converting 151 grams to moles, so we multiply by this value.
[tex]151 \ g \ SO_2 *\frac {64.068 \ g \ SO_2} {1 \ mol \ SO_2}[/tex]
Flip the ratio so the units of grams of sulfur dioxide cancel.
[tex]151 \ g \ SO_2 *\frac {1 \ mol \ SO_2}{64.068 \ g \ SO_2}[/tex]
[tex]151 *\frac {1 \ mol \ SO_2}{64.068 }[/tex]
[tex]\frac {151}{64.068 } \ mol \ SO_2[/tex]
[tex]2.356870825 \ mol \ SO_2[/tex]
3. RoundThe original measurement of grams (151) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 6 in the thousandth place tells us to round the 5 in the hundredth up to a 6.
[tex]2.36 \ mol \ SO_2[/tex]
151 grams of sulfur dioxide is approximately 2.36 moles of sulfur dioxide.