Consider the reaction of Copper with Sulfur: Cu +S → Cus. Using the
below data, which of following is true?

A. Sulfur was the limiting reactant; 1.61 g of sulfur were reacted.
B. Copper was the limiting reactant; 1.32 g of sulfur were unreacted
C. Copper (II) sulfide was the limiting reactant; 4.78 g was formed.
D. There was no limiting reactant; both reactants remained in excess.

Answer:

cornea,Jens and pupil

tma Po yan

Answer:

What's the equation?

Answer:

14.5 g

Explanation:

its true

Answer:

59.3g

Explanation:

Mole = no. Molecules/6.02×10^23

Mole = (8.12×10^23)/(6.02×10^23)

Mole = 1.35mole

Molar mass of CO2 = 12+ 2(16)

Molar mass= 12 + 32= 44g/mol

Mole = mass/molar mass

Mass = Mole × molar mass

Mass = 1.35× 44

Mass= 59.35g

Answer:

The Cell Wall

Explanation:

Answer:

[tex][H^+]=[OH^-]=1.0x10^{-7}M[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the relative concentrations by using the following equations:

[tex]pH+pOH=14\\\\pH=-log([H^+])\\\\pOH=-log([OH^-])[/tex]

Thus, we realize that both pH and pOH are the same, 7, so that the concentration of H⁺ and OH⁻ are virtually the same, and can be calculated as follows:

[tex][H^+]=[OH^-]=10^{-7}=1.0x10^{-7}M[/tex]

Regards!

Answer:

Mass of the ball is . 42 kg. The density of air is 1.2 kg/m3.

Explanation:

Answer:

ee = 96%

Explanation:

Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the  differences of amounts of enantiomers and the total amunt. that is:

ee = |A-B|/ A+B * 100

ee = |98%-2%| / 98+2 * 100

Answer:

16.48 mL.

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 1 atm

Initial temperature (T₁) = 0 °C

Initial volume (V₁) = 30 mL

Final temperature (T₂) = 27 ° C

Final pressure (P₂) = 2 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 0 °C

Initial temperature (T₁) = 0 °C + 273

Initial temperature (T₁) = 273 K

Final temperature (T₂) = 27 °C

Final temperature (T₂) = 27 °C + 273

Final temperature (T₂) = 300 K

Finally, we shall determine the final volume of the gas. This can be obtained as:

Initial pressure (P₁) = 1 atm

Initial temperature (T₁) = 273 K

Initial volume (V₁) = 30 mL

Final temperature (T₂) = 300 K

Final pressure (P₂) = 2 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

1 × 30 / 273 = 2 × V₂ / 300

30 / 273 = 2 × V₂ / 300

Cross multiply

273 × 2 × V₂ = 30 × 300

546 × V₂ = 9000

Divide both side by 546

V₂ = 9000 / 546

V₂ = 16.48 mL

Therefore, the final volume of the gas is 16.48 mL.

Answer:

C

Explanation:

Answer:

True

Explanation:

hope this helps mate

Answer:

Ang kasaysayan ng mga laro ay nagmula sa sinaunang nakaraan ng tao. Ang mga laro ay isang mahalagang bahagi ng lahat ng mga kultura at isa sa pinakamatandang anyo ng pakikipag-ugnay sa lipunan ng tao. Ang mga laro ay gawing pormal na pagpapahayag ng paglalaro na nagpapahintulot sa mga tao na lumampas sa agarang imahinasyon at magdirekta ng pisikal na aktibidad. Kasama sa mga karaniwang tampok ng mga laro ang kawalan ng katiyakan sa kinalabasan, napagkasunduang panuntunan, kumpetisyon, magkakahiwalay na lugar at oras, mga elemento ng kathang-isip, mga elemento ng pagkakataon, iniresetang mga layunin at personal na kasiyahan.

Nakukuha ng mga laro ang mga ideya at pananaw sa mundo ng kanilang mga kultura at ipinapasa sa hinaharap na henerasyon. Ang mga laro ay mahalaga bilang mga kaganapan sa pagbubuklod ng kultura at panlipunan, bilang mga tool sa pagtuturo at bilang mga marka ng katayuan sa lipunan. Bilang pampalipas ng oras ng pagkahari at ng mga piling tao, ang ilang mga laro ay naging karaniwang tampok ng kultura ng korte at binigyan din bilang mga regalo. Ang mga larong tulad ni Senet at ang Mesoamerican ball game ay madalas na pinuno ng mitiko at ritwal na relihiyosong kahalagahan. Ang mga larong tulad ng Gyan chauper at The Mansion of Happiness ay ginamit upang magturo ng mga aralin na pang-espiritwal at etikal habang sina Shatranj at Wéiqí (Go) ay nakita bilang isang paraan upang mapaunlad ang madiskarteng pag-iisip at kasanayan sa kaisipan ng mga elite sa politika at militar.

Answer:

A mixture is matter that can vary in composition. ... A heterogeneous mixture is a type of mixture in which the individual substances are not evenly mixed. Because the substances in a heterogeneous mixture are not evenly mixed, two samples of the same mixture can have different amounts of the substance.

Explanation:

Answer:

[tex]pH = - log(7.6 \times {10}^{ - 5} ) \\ pH = 4.12[/tex]

answer is 2546 h

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Answer:

The density should be 2.8 g/mL.

Explanation:

To solve this problem, we must remember that we find density using the following formula:

Density = mass/volume

Therefore, if we plug in the values we are given in the problem, we get:

Density = 70 g / 25 mL

Density = 2.8 g/mL

Therefore, the correct answer is 2.8 g/mL.

Hope this helps!

Answer:

-713 kJ

Explanation:

Step 1: Write the balaned thermochemical equation

4 NH₃(g) + 3 O₂(g) → 2 N₂(g) + 6 H₂O(g)  ΔΗ = -1267 kJ

Step 2: Calculate the moles corresponding to 38.4 g of NH₃

The molar mass of NH₃ is 17.03 g/mol.

38.4 g × 1 mol/17.03 g = 2.25 mol

Step 3: Calculate the  enthalpy change to burn 2.25 mol of ammonia

According to the thermochemical equation, 1267 kJ are released per 4 moles of ammonia that react.

2.25 mol × (-1267 kJ/4 mol) = -713 kJ

Answer:

128.4 g

Explanation:

Step 1: Given data

Step 2: Determine the total mass of the sample

The mass percent of element X in X(NO₃)₂ is 52.55%, that is, there are 52.55 g of X every 100 g of X(NO₃)₂. Then, the mass of X(NO₃)₂ that contains 67.50 g of X is:

67.50 g X × 100 g X(NO₃)₂/52.55 g X = 128.4 g X(NO₃)₂

Answer:

V1 = 8.3 mL

Explanation:

Step 1: Given data

Initial volume (V1): ?

Initial concentration (M1): 6.0 M

Final volume (V2): 50.0 mL

Final concentration (M2): 1.0 M

Step 2: Calculate the volume of the initial solution

We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the initial solution using the dilution rule.

M1 × V1 = M2 × V2

V1 = M2 × V2 / M1

V1 = 1.0 M × 50.0 mL / 6.0 M = 8.3 mL

We will take 8.3 mL of the 6.0 M solution and add water until we have 50.0 mL.

Answer:

1

Explanation:

hhihh2

Answer:

Explanation:

Half life T is 432 years .

432 years = 432 x 365 x 24 x 60 x 60 s

= 1.36 x 10¹⁰ s .

No of atoms in 5.54 g of sample of Am N= 6.02 x 10²³ x 5.54 / 241

N = 138.38 x 10²⁰ atoms .

disintegration constant λ = .693 / half life

λ =  .693 / 1.36 x 10¹⁰ s .

= .51 x 10⁻¹⁰ .

dN / dt = λN

= .51 x 10⁻¹⁰  x  138.38 x 10²⁰

= 70.57 x 10¹⁰ per second.

No of particles emitted = no of disintegration per second

No of particles emitted =  70.57 x 10¹⁰ per second.

Answer:

CH3CH2CH2CH3 + Cl2 --------> CH3CH2CH2CH2Cl + HCl

Explanation:

Alkanes react with halogens in the presence of light to yield alkyl halides. The degree of substitution increases as the reaction progresses. The reaction occurs by free radical mechanism.

The reaction between butane and chlorine molecule to yields a monosubstitution product occurs as follows;

CH3CH2CH2CH3 + Cl2 --------> CH3CH2CH2CH2Cl + HCl

Answer:

C. It was based on the thoughts of an early philosopher.

Answer:

Explanation:

25.71 × 10²³ molecules are there in 4.27 mol of tungsten (VI) oxide, [tex]WO_3[/tex].

To find the Number of molecules of substance multiply the number of moles of given substance by Avogadro number

Number of molecules = Number of moles × Avogadro's Number

Avogadro's number [tex](N_{A})[/tex] is the number of particles in one mole of substance. 6.023 × 10²³ is known as Avogadro's constant / Avogadro's number.

Avogadro's number = 6.023 × 10²³ [tex]mol^{-}[/tex]

Now,

Number of molecules in 4.27 moles of [tex]WO_{3}[/tex]

= Number of moles of [tex]WO_{3}[/tex] × Avogadro's Number

= 4.27 × 6.023 × 10²³

= 25.71 × 10²³ molecules

Thus, we can say that 25.71 × 10²³ molecules are there in 4.27 moles of tungsten (VI) oxide, [tex]WO_3[/tex].

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The symbol of the element in which it's ion has 38 protons, 36 electrons, and 40 neutrons is ⁷⁸₃₈Sr

To obtain the symbol of the element, we shall obtain the mass number of the element. This is shown below:

Mass number = Proton + Neutron

Mass number = 38 + 40

Mass number = 78

Finally, we shall determine the symbol of the element. Details below:

From the periodic table, the element with atomic number of 38 is Strontium.

Thus, the symbol of the element will be ⁷⁸₃₈Sr

Learn more about symbol:

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Answer: The number of electrons in given amount of silver are [tex]1.839\times 10^{24}[/tex]

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

......(1)

We are given:

Mass of silver = 7.0 g

Molar mass of silver = 107.87 g/mol

Putting values in equation 1:

[tex]\text{Moles of silver}=\frac{7.0g}{107.87g/mol}=0.0650 mol[/tex]

According to the mole concept:

1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of particles.

Number of electrons per atom of silver = 47

So, 0.065 moles of silver will contain = [tex](0.065\times 6.022\times 10^{23}\times 47)=1.839\times 10^{24}[/tex] number of electrons

Hence, the number of electrons in given amount of silver are [tex]1.839\times 10^{24}[/tex]

Answer:

See explanation

Explanation:

Let us look at the reaction again;

Cr2O7 2- (aq) + H2O(l)⇄ 2CrO4 2-(aq) + 2H^+(aq)

When we add sodium hydroxide to the system as shown, the hydroxide ion removes the hydrogen ion thereby leaving a large concentration CrO4^2-(aq) in the system this causes the solution to turn green(equilibrium position shifts to the right).

The net ionic equation is;

OH^-(aq) + H^+(aq) ----> H2O(l)

The reaction;

OH^-(aq) + H^+(aq) ----> H2O(l) is exothermic hence, if the temperature of the system is increased, the equilibrium position will shift towards the left hand side and the solution turns orange.