A solution is prepared by dissolving 74.2 g NaI in 527 g of water. Determine the mole fraction of NaI if the final volume of the solution is 565 mL.

a. 8.84 times 10^-2
b. 1.40 times 10^-3
c. 6.04 times 10^-3
d. 1.32 times 10^-2
e. 1.65 times 10^-2

Answer:

2LiCl + 3O2  

Explanation:

Answer:

(C) 2LiCl + 3O2

Explanation:

sub to pewds

Answer:

A.

............

Explanation:

Ductility

Answer:

49.4 cal

Explanation:

Step 1: Given data

Composition of 1 cup of watermelon: 0.9 g protein, 0.2 g fat, and 11 g carbohydrate.

Step 2: Calculate the calories provided by 0.9 g of protein

1 g of protein yields 4 cal.

0.9 g × (4 cal/1 g) = 3.6 cal

Step 3: Calculate the calories provided by 0.2 g of fat

1 g of fat yields 9 cal.

0.2 g × (9 cal/1 g) = 1.8 cal

Step 4: Calculate the calories provided by 11 g of carbohydrate

1 g of carbohydrate yields 4 cal.

11 g × (4 cal/1 g) = 44 cal

Step 5: Calculate the total number of calories

3.6 cal + 1.8 cal + 44 cal = 49.4 cal

Answer:

b

Explanation:

b

Answer:

B- heat from an oven

Explanation:

temperature isn’t energy

Answer:

you are correct, the answer is 2 g

Answer:

Explanation:

The acceleration of gravity of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

f is the force

m is the mass

From the question we have

[tex]a = \frac{800}{75} = \frac{32}{3} \\ = 10.666666...[/tex]

We have the final answer as

Hope this helps you

Answer:

F2

Explanation:

F2 is the only element with only 1 element and eveything else is formed by 2 non-metals

Answer:d I believe

Explanation:

Answer:

5.3 mol CO₂

General Formulas and Concepts:

Chemistry - Stoichiometry

Explanation:

Step 1: Define

3.2 × 10²⁴ molecules CO₂

Step 2: Convert

[tex]3.2 \cdot 10^{24} \ mc \ CO_2(\frac{1 \ mol \ CO_2}{6.022 \cdot 10^{23} \ mc \ CO_2} )[/tex] = 5.31385 mol CO₂

Step 3: Check

We are given 2 sig figs. Follow sig fig rules.

5.31385 mol CO₂ ≈ 5.3 mol CO₂

Answer:

The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.

Explanation:

You know  the balanced reaction:

4 NA + O₂ ⟶ 2 Na₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:

Being:

the molar mass of the compounds participating in the reaction is:

Then by stoichiometry of the reaction they react and are produced:

Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?

[tex]massofNa_{2}O =\frac{4 grams of Na*124 gramsofNa_{2}O }{92 grams of Na}[/tex]

mass of Na₂O=5.39 g

The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.

Answer:

Organelles allow for various functions to occur in the cell at the same time. Despite their fundamental similarities, there are some striking differences between animal and plant cells (see Figure 1). Animal cells have centrosomes (or a pair of centrioles), and lysosomes, whereas plant cells do not.

Explanation:

Answer:

The answer is "152.28%".

Explanation:

Formula for ideal gas:

PV=nRT

equation:

[tex]\to (P+\frac{an^2}{v^2})(v-nb)=nRT \\\\a=2.253 \frac{L^2 atm}{mol^2}\\\\b=4.278 \times 10^{-2} \frac{L}{mol}\\\\n=10.68 \ mol\\\\v= 0.8295 \ L\\\\\to (P+\frac{an^2}{v^2})= \frac{nRT}{(v-nb)}\\\\\to (P)= \frac{nRT}{(v-nb)} - \frac{an^2}{v^2} \\\\[/tex]

          [tex]= \frac{10.68 \times 0.0082 \times 501.9}{0.8295 - (10.68 \times 4.278 \times 10^{-2})} - \frac{2.253 \times 10.68^2}{0.8295^2}\\\\=\frac{43.9543944}{0.4568904} - \frac{256.982587}{0.68807025}\\\\ = 96.2033661- 373.483067\\\\=-277.279701 \\\\ =-277.279701 \ atm[/tex]

The pressure of gas:

[tex]= (530.3 + 277.279701) \ atm\\\\= 807.579701 \ atm \\\\[/tex]

Calculating the pressure percentage:

[tex]=\frac{807.579701}{530.3} \times 100\\\\= 152.28 \%[/tex]

Answer:

Electronegativity  = 1.87.

Ionic radius = 109 pm.

Atomic radius = -39 pm

First ionization energy = 410 kJ/mol

Explanation:

Hello!

In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.

In such a way, the required comparison is written below:

Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.

Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.

Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm

First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol

It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.

Best regards!

Oxygen and Beryllium are the elements that pertain to the same period and different groups 6A and 2A.

Ionic radius, atomic radius, electronegativity and first ionization energy are periodic qualities that have selective trends.

Differences in the trends are:

This can be explained as:

= 140 pm (oxygen)- 31 pm (beryllium)

= 109.

= 73 pm (oxygen) - 112 pm (beryllium)

= -39 pm

= 3.44 (oxygen) - 1.57 (beryllium)

= 1.87

= 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium)

= 410 kJ/mol

Therefore, these data tell that ionic radius, electronegativity and first ionization energy increases from left to right in a period whereas, the atomic radius increases from right to left in a period.

To learn more about periodic trends follow the link:

https://brainly.com/question/12074167

Answer:

True

Explanation:

Answer:

The major jet streams on Earth are westerly winds that flow west to east.

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

Answer:

NaBr + Cl2 = NaCl + Br2 - Chemical Equation Balancer.

Explanation:

Answer:

Chemical Equation: (NH4)2S(aq) + 2KOH(aq) -> K2S(aq) + 2NH4OH(s)

Complete Ionic Equation: 2NH4^+(aq) + S^2-(aq) + 2K^+(aq) 2OH^-(aq) -> 2K^+(aq) + S^2- (aq) + 2NH4OH(s) 2NH4^+(aq) + OH^-(aq) -> NH4OH(s)

Explanation:

Ammonium sulfate and potassium hydroxide interact in this process to produce aqueous potassium sulfate, gaseous ammonia, and liquid water as byproducts. In essence, an acid-base reaction is taking place here, with the ammonium ion contributing a hydrogen ion to the potassium hydroxide's hydroxide ion.

Chemical formulae and symbols are used to express a chemical reaction in a chemical equation.

A chemical equation represents a chemical reaction using symbols and the chemical formulas of the reactants and products involved. In this process, ammonia and hydrochloric acid are reacting to generate ammonium chloride. For solids, the symbol is (s).

( NH₄ ) 2S ₍ aq ₎ + 2KOH ₍ aq ₎ ⇒ K₂S ₍ aq ₎  + 2NH₄

Thus, Aqueous solutions of ammonium sulfide and potassium hydroxide are mixed to formed potassium sulphide and ammonia.

To learn more about chemical equation follow the link below;

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#SPJ3

Answer:

In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.

Answer:

7.5 x 10^14 Hz

Explanation:

Answer:

565 kJ

Explanation:

[tex]H_v[/tex] = Heat of vaporization of water at 100°C = 2260 J/g

m = Mass of water = 250 g

Heat required is given by

[tex]Q=mH_v[/tex]

[tex]\Rightarrow Q=250\times 2260[/tex]

[tex]\Rightarrow Q=565000\ \text{J}[/tex]

[tex]\Rightarrow Q=565\times 10^3\ \text{J}[/tex]

[tex]\Rightarrow Q=565\ \text{kJ}[/tex]

The heat required to vaporize the water is 565 kJ.

Explanation:

The number of orbitals in the sublevels are given below:

     Sublevels                   Orbitals

        s                                  1

        p                                 3

        d                                 5

        f                                   7

a. ls       -  1 orbital

b. 5s      - 1 orbital

c. 4d     - 5 orbitals

d. 4f       - 7 orbitals

e. 7s       - 1 orbital

f. 3p       - 3 orbitals

g. Entire 5th principal energy level : for s, p ,  f  

          1 + 3 + 5 + 7 + 9  = 25

h. 6d   - 5

Answer: 0.006 km

Explanation:

divide the length value by 1000

so 6/1000 = 0.006

The specific heat of metal = 0.897 J/g° C  

Heat absorbed can be calculated using the formula:  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q = 8.793 J

m = 2 g

∆t = 4.9 °C

[tex]\tt c=\dfrac{Q}{m.\Delta t}\\\\c=\dfrac{8.793}{2\times 4.9}\\\\c=0.897~J/g^oC[/tex]

Answer:

Q1 = C * m * dT

Q2 = Qm * m

Qtotal = Q1 + Q2

Q1 - is amount of energy you need to apply to heat oxygen from the current temperature till you reach the melting temperature. Only if the oxygen is below to melting temperature.

C - is calorific capacity of oxygen -- better look at tables, it is a constant value

m - is the amount of oxygen, we will use moles because the other data shows moles, but could be grams, kg, etc.

dT - is the diference of temperatures between the current and the melting one. The melting temperature is constant and you can find it on tables, then (Tm - To)

Q2 is the amount of energy you have to add to melt oxygen once the oxygen has reached the melting temperature (Tm)

Qm is a constant value you could find on tables, depends on the mass of oxygen and is due to internal processes as changes in atomic distributions

If the oxygen is initially at melting temperature (melting point) you only need to know Q2, as dT = 0

I will do an example for you, but in future you should provide data of constants, it takes very long to find them in books or internet.

Data from tables

Tm =  54.36 K

C = 29.378 J/mol K this is at 25 C (or 298 K), is not really correct, you should look at its value at less than 54.36 K, but you can use it here.

Qm = 0.444 kJ/mol

Problem -- you have 44.33g of Oxygen -- Molecular weight of O2 is 32 g/mol

So you have 44.33/32 = 1.385 moles of oxygen

a) if oxygen is already at melting temperature: you only have to melt it

Qtotal = Q1 + Q2 = [0 (dT = 0) + Qm * m] = 0.444 * 1.385 = 0.615 kJ = 615 J

b) supposing an initial temperture of 50 K: now you have to heat oxygen till melting temperature and then melt it.

Q1 = C * m * dT = 29.378 * 1.385 * (54.36 - 50) = 177.442 J

Q2 = Qm * m = 615 J

Qtotal = 177.442 + 615 = 792.44 J

Explanation:

Answer:

In this case, 1 mole of liquid oxygen requires 3.41 kJ of heat in order to perform that liquid → solid. your sample from grams to moles by using oxygen's molar mass.

Answer:

d. IF3

Explanation:

The Octet rule posits that atoms gain, atom lose, or share electrons in order to have a full valence shell of 8 electrons. This statement occurs when atoms also combine to form molecules until they attain or share eight valence electrons either by losing or gaining eletrons.

From the given options, a valid Lewis structure that cannot be drawn without violating the octet rule is IF3

Answer:

Volume V2 = 948.13 ml

Explanation:

Given:

Volume V1 = 525 ml

Temperature T1 = -25°C + 273.15

Volume V2 = ?

Temperature T1 = 175°C + 273.15

Computation:

V1 / T1 = V2 / T2

525 / [-25°C + 273.15] = V2 / [175°C + 273.15]

Volume V2 = 948.13 ml

The new volume, in milliliters, of the gas is 948 L

From the question,

We are to determine the new volume of the gas.

From Charles' law which states that the volume of a fixed mass of gas is directly proportional to the temperature (in Kelvin) provided that the pressure remains constant.

That is,

V ∝ T

Then,

V = kT

Therefore, we can write that

[tex]\frac{V_{1} }{T_{1}} = \frac{V_{2}}{T_{2}}[/tex]

Where [tex]V_{1}[/tex] is the initial volume

[tex]T_{1}[/tex] is the initial temperature

[tex]V_{2}[/tex] is the final volume

and [tex]T_{2}[/tex] is the final temperature

From the given information,

[tex]V_{1} = 525 \ mL[/tex]

[tex]T_{1} = -25 ^{\circ}C = -25 + 273.15 \ K = 248.15 \ K[/tex]

[tex]T_{2} = 175 ^{\circ} C = 175 +273.15 \ K =448.15 \ K[/tex]

Putting the values into the formula, we get

[tex]\frac{525}{248.15} = \frac{V_{2} }{448.15}[/tex]

∴ [tex]V_{2} = \frac{525 \times 448.15}{248.15}[/tex]

[tex]V_{2} = \frac{235278.75}{248.15}[/tex]

[tex]V_{2} = 948.13 \ L[/tex]

V₂ ≅ 948 L

Hence, the new volume, in milliliters, of the gas is 948 L

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Heat required : 1.523 kJ

Reaction

N₂+2O₂⇒2NO₂ AH = 67.7 kJ

mol of N₂:(use Pv=nRT)

[tex]\tt n=\dfrac{3.32\times 0.304}{0.082\times 373}=0.033[/tex]

mol of O₂:

[tex]\tt n=\dfrac{3.32\times 0.41}{0.082\times 373}=0.045[/tex]

Limiting reactant :

N₂ : O₂

[tex]\tt \dfrac{0.033}{1}\div \dfrac{0.045}{2}=0.033:0.0225[/tex]

Limiting reactant : O₂(smaller ratio)

mol NO₂ = mol O₂ = 0.045

2 mol ⇒ 67.7 kJ, so for 0.045 mol, heat required :

[tex]\tt \dfrac{0.045}{2}\times 67.7=1.523~kJ[/tex]

The heat required to produce the maximum yield of NO2 is 1.52 kJ.

The balanced reaction equation is;

N2(g) + 2 O2(g) -> 2 NO2(g)

The number of moles of each gas is obtained from the ideal gas equation as follows;

For N2;

P = 3.32 atm

V = 304 ml or 0.304 L

T = 100ºC + 273 = 373 K

n = ?

R = 0.082 atmLK-1mol-1

From;

PV = nRT

n = PV/RT

n = 3.32 atm × 0.304 L/0.082 atmLK-1mol-1 × 373 K

n = 0.033 moles

For O2;

PV = nRT

n =  PV/RT

V = 410 ml or 0.410 L

P = 3.32 atm

R = 0.082 atmLK-1mol-1

T = 100°C + 273 = 373 K

n =  3.32 atm × 0.410 L/0.082 atmLK-1mol-1 × 373 K

n = 0.045 moles

1 mole of N2 reacts with 2 moles of O2

0.033 moles of N2 reacts with 0.033 moles × 2 moles/1 mole

= 0.066 moles

2 moles of O2 yields 2 moles of NO2

0.045 moles O2 yields 0.045 moles × 2 moles/2 moles = 0.045 moles

If 2 mole of NO2 requires 67.7 kJ

0.045 moles of NO2 requires 0.045 moles  × 67.7 kJ/2 mole

= 1.52 kJ

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