what is the difference between red phosphorus and white phosphorus?​

Answer:

hydrogen bonding

Explanation:

just took the test :D

[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]

[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]

= 2 × 23 + 2 × 52 + 2 × 16

= 182 grams

1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g

8 moles weigh = 8× 182

=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]

or

[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]

Answer:

C) METALLIC IS THE CORRECT ANSWER

Explanation: I just did the exam

Answer:

4.4 mL

Explanation:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed?

Step 1: Write the balanced equation

CH₄(g) + 4 Cl₂(g) ⇒ 4 HCl(g) + CCl₄(g)

Step 2: Establish the appropriate volume ratio

For gases under the same conditions, the volume ratio is equal to the molar ratio. The molar ratio of CH₄ to HCl is 1:4.

Step 3: Calculate the volume of HCl produced from 1.1 mL of CH₄

1.1 mL CH₄ × 4 mL HCl/1 mL CH₄ = 4.4 mL HCl

Given question is:

Given a food with a relative humidity of 81%, what is the water activity in that food?

Explanation:

Water activity in food can be determined by using the below-shown formula:

[tex]water activity=\frac{equilibrium relative humidity}{100}[/tex]

Equilibrium is established between the vapor pressure of food and the surrounding air media.

Thus, for the given food the relative humidity is 81%.

hence, its water activity is

[tex]81/100\\=0.81[/tex]

Answer:

O2 is a covalent substance while NaCl is an ionic substance

Explanation:

In O2 molecule, the bond is between 2 oxygen atoms which are non - metals. Thus, this is a covalent bond since it involves 2 non metals.

Whereas, for the NaCl molecule, the bond is between a metal sodium (Na) and a non metal Chloride(Cl) and thus we can say this is an ionic bond.

Thus the difference is that O2 is a covalent substance while NaCl is an ionic substance.

Answer:

when naming ionic compounds — those are only used in naming covalent molecular compounds. Do NOT use prefixes to indicate how many of each element is present; this information is implied in the name of the compound. since iron can form more than one charge. Ionic Compounds Containing a Metal and a Polyatomic Ion.

Answer:

the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Explanation:

Given the data in the question;

Co = 53 or [ 53 wt% B-47 wt% A ]

W∝ = 0.5 = Wβ

Cβ = 92 or [ 92 wt% B-8 wt% A ]

Now, lets set up the Lever rule for W∝ as follows;

W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]

so we substitute our given values into the expression;

0.5 = [ 92 - 53 ] / [ 92 - C∝ ]

0.5 = 39 /  [ 92 - C∝ ]

0.5[ 92 - C∝ ] = 39

46 - 0.5C∝  = 39

0.5C∝ = 46 - 39

0.5C∝ = 7

C∝ = 7 / 0.5

C∝ = 14  or [ 14 wt% B-86 wt% A ]

Therefore, the composition of the ∝ phase C∝ = 14  or [ 14 wt% B-86 wt% A ]

Answer:

gravity, ground, friction, rolling resistance, and air resistance.

Answer:

The Rf value change will be  >  0.5

Explanation:

Given that Change in Solvent is proportional to change in polarity of solvent system

The change from solvent composition of 60/40  to 30/70 will cause an increase in the polarity of the system .

and Increase in Polarity = Increase in Rf value  because the compound will move to a higher distance

Answer:

Molar mass for eugenol is 161.3 g/mol

Explanation:

This question talks about freezing point depression:

Our solute is eugenol.

Our solvent is camphor.

Formula to state the freezing point depression difference is:

ΔT = Kf . m . i where

ΔT = Freezing T° of pure solvent - Freezing T° of solution

In this case ΔT = 0.62°C

Kf for camphor is: 37°C /m

As eugenol is an organic compund, i = 1. No ions are formed.

To state the molar mass, we need m (molal)

Molal are the moles of solute in 1kg of solvent. Let's replace data:

0.62°C = 40 °C/m . m . 1

0.62°C / 40 m/°C = 0.0155 m

We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg

0.0155 molal = moles of solute / 0.1 kg

0.0155 m/kg . 0.1 kg = 0.00155 moles

We know that these moles are contained in 250 mg, so the molar mass will be:

0.25 g / 0.00155 mol = 161.3 g/mol

Notice, we convert mg to g, for the units!

QUESTION:- Which particle has a mass of 9.11 x 10^-28g and charge of -1?

OPTIONS:-

A. electron

B. proton

C. neutron

ANSWER:-

CHARGE ON PROTRON IS +1 AND IT HAS MASS OF [tex]1.6 \times 10 {}^{ - 27} [/tex] SO IT CANNOT BE URE ANSWER

THERE IS NO CHARGE ON NEUTRON AND HAS MASS ALMOST EQUAL TO THE PROTON SO IT ALSO CANNOT BE URE ANSWER

MASS OF THE ELECTRON:- [tex]9.11 \times 10^{ - 28} [/tex]

CHARGE ON ELECTRON:- [tex] -1[/tex]

SO URE ANSWER IS ELECTRON

Answer:

% yield of reaction is 26.4

Explanation:

The reaction is:

Mg + I₂ →  MgI₂

Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:

10 g . 1mol / 24.3 g = 0.411 moles of Mg

95 g . 1mol / 253.8g = 0.374 moles of I₂

Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine

For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.

As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂

We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g

To calculate the percent yield:

% yield = (yield produced /theoretical yield) . 100

% yield = (27.5 g/ 104g) . 100 = 26.4

Answer:

KOH + HClO4 = K(ClO4) + H2O

Explanation:

KClO4 + H2O : mix and match the letters if you need to guess

Explanation:

Iron has 2 atoms and 3atoms.

So,X=2,3

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]

[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]

[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]

By rearrangement:

[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]

Multiplying through  by (-) and solving the quadratic equation:

[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]

[tex](-0.00015 + x) (0.000168 + x) =0[/tex]

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]

= 10.71%

Answer:

209

83 Bi

Explanation:

213 213 - 4 4

85 At = 85 - 2 Y + 2 He

Solution :

[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]

Explanation 1 :

Spontaneity of the reaction is based on two factors :

-- the tendency to acquire a state of minimum energy

-- the energy of a system to acquire a maximum randomness.

Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.

Explanation 2 :

A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].

It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.

Answer:

[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

Answer:

1,3-dibromobenzene

Explanation:

An image of the compound described in the question is attached to this answer.

We need to reiterate here the rules of IUPAC nomenclature. The substituents in a compound must be named in such a way that they have the lowest number.

The compound described may also be named as 1,5-dibromobenzene but this name is disallowed because it gives the substituents a higher number than 1,3-dibromobenzene.

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

Answer:

B

Explanation:

liquids is produce therefore it will have the less Entropy

Answer:

B.

Explanation:

if the # of molecules (of gas) on the product side is less than # of molecules on the reactant side = entropy is decreasing, and vice versa.

if the # of molecules on the reactant side is less than # of molecules on the product side = entropy is increasing (it is more disorderly, chaotic)

A. 2 → 2 + 1 entropy increases

B. 1 + 4 → 1 + 1 entropy decreases

C. 2 → 1 + 3 entropy increases

D. 2 + 1 → 2 + 2 entropy increases

Answer:

1) 0.00498 mol Cu.

2) 0.00000374 mol CO₂

Explanation:

Question 1)

We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.

Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:

[tex]\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}[/tex]

We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:

[tex]\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}[/tex]

Cancel like terms:

[tex]=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }[/tex]

Simplify:

[tex]\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}[/tex]

Use a calculator:

[tex]= 0.004981... \text{ mol Cu}[/tex]

Since the resulting answer must have three significant figures:

[tex]= 0.00498\text{ mol Cu}[/tex]

So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.

Question 2)

We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.

By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:

[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}[/tex]

To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:

[tex]\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}[/tex]

Cancel like terms:

[tex]\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}[/tex]

Simplify:

[tex]\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}[/tex]

Use a calculator:

[tex]=0.000003736...\text{ mol CO$_2$}[/tex]

Since the resulting answer must have three significant figures:

[tex]= 0.00000374\text{ mol CO$_2$}[/tex]

So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.

Answer:

Explanation:

by definition, 1 mole contains 6.02 x 10^23 of atoms (for elements) or molecules (for compounds)

3.00 x 10^21 atoms of copper / 6.02 x 10^23 of atoms

= 0.004983 moles of copper

= 4.98 x 10^(-3) moles of copper

2.25 x 10^18 molecules of carbon dioxide / 6.02 x 10^23 of molecules

= 0.000003737 moles of carbon dioxide

= 3.74 x 10^(-6) moles of carbon dioxide

Answer:

1.) Propanone (ketone)

2.) Ethanal( aldehyde)

3.) 3-phenyl-2-propenal (aldehyde)

4.) Butanone (ketone)

5.) Ethanol ( alcohol)

6.) 2-propanol (alcohol)

Explanation:

In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.

Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.

Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal

Answer:

i think this is a biology question but regardless a threatened species is same as endangered which is going to become extinct in the near future but i say the best answer is threatened with becoming extinct in the near future.

Answer:

Density of mercury is 13600 kg

Answer:

80g

Explanation:

mass oxygen present in 1 mole of p4010

16×10=160gm

similarly

for 0.5 moles of p4010 160/2= 80gm

The number of atoms present in 0.45 moles of P₄O₁₀ is 1.08 x 10²³ atoms.

To determine the number of atoms, we use Avogadro's number, which states that there are approximately 6.022 x 10²³ particles (atoms, molecules, or formula units) in one mole of a substance.

In this case, we are given 0.45 moles of P₄O₁₀. To calculate the number of atoms, we multiply the number of moles by Avogadro's number:

Number of atoms = 0.45 moles P₄O₁₀ x (6.022 x 10²³ atoms / 1 mole)

Number of atoms = 2.7139 x 10²³ atoms

Rounding to three significant figures, the number of atoms present in 0.45 moles of P₄O₁₀ is approximately 1.08 x 10²³ atoms.

To learn more about atoms   here

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