CHEMISTRY COLLEGE

Balance P4O10 + H2O → H3PO4

Answers

Answer 1

Answer:

Answer:

P4010 + 6H20 → 4H3PO4

Explanation:

i rlly h8 this lesson but idt chem anymore but i used to do these alot for the last past 2 yrs

p-4 p-1

o-11 o-4

h-2 h-3

(these r for me so dont rlly mind it)

hope it helped tho:)))

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When a chemical reaction occurs blank happens

Answers

Answer:

In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

Explanation:

Need help I don't remember what to do?

Answers

I'm just doing the ones that you don't have numbers already for.
2.) just leave it alone and it's correct
3.) Mg + 2AgNo3 --> Mg(No3)2 + 2Ag
5.) just leave it alone and it's correct
8.) 10C3H8O3 + 15O2 --> 30CO2 + 4H2O
10.) P4 + 6Br2 --> 4PBr3
12.) 2FeCl3 + 6NaOH --> 2Fe(OH)3 + 6NaCl
13.) 2CH3OH + 3O2 --> 2CO2 + 4H2O
14.) 2Al + 3Cu(NO3)2 --> 2Al(NO3)3 + 3Cu
15.) 3CaCl2 + 2K3AsO4 --> Ca3(AsO4)2 + 6KCl
16.) 2NH3 --> N2 + 3H2
17.) 2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
19.) Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
I hope this helps you!!

A cubic box with sides of 20.0 cm contains 2.00 × 1023 molecules of helium with a root-mean-square speed (thermal speed) of 200 m/s. The mass of a helium molecule is 3.40 × 10-27 kg. What is the average pressure exerted by the molecules on the walls of the container? (The Boltzmann constant is 1.38 × 10-23 J/K and the ideal gas constant is R = 8.314 J/mol•K .) (12 pts.)

Answers

Answer:

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Explanation:

Side of the cubic box = s = 20.0 cm

Volume of the box ,V= $s^3$

$V=(20.0 cm)^3=8000 cm^3=8* 10^(-3) m^3$

Root mean square speed of the of helium molecule : 200m/s

The formula used for root mean square speed is:

$\mu=\sqrt{(3kN_AT)/(M)}$

where,

= root mean square speed

k = Boltzmann’s constant = $1.38* 10^(-23)J/K$

T = temperature = 370 K

M = mass helium = $3.40* 10^(-27)kg/mole$

$N_A$ = Avogadro’s number = $6.022* 10^(23)mol^(-1)$

$T=(\mu _(rms)^2* M)/(3kN_A)$

Moles of helium gas = n

Number of helium molecules = N = $2.00* 10^(23)$

N = $N_A* n$

Ideal gas equation:

PV = nRT

Substitution of values of T and n from above :

$PV=(N)/(N_A)* R* (\mu _(rms)^2* M)/(3kN_A)$

$PV=(N* R* \mu ^2* M)/(3k* (N_A)^2)$

$R=k* N_A$

$PV=(N* \mu ^2* M)/(3)$

$P=(2.00* 10^(23)* (200 m/s)^2* 3.40* 10^(-27) kg/mol)/(3* 8* 10^(-3) m^3)$

$P=1133.33 Pa =1.133 kPa$

(1 Pa = 0.001 kPa)

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Final answer:

The question asks for the average pressure exerted by helium gas molecules on the walls of a cubic container. Using the equation PV = Nmv^2, we can calculate pressure by substituting the given values for volume, number of molecules, mass of one molecule, and root-mean-square speed.

Explanation:

The question is asking to calculate the average pressure exerted by helium gas molecules on the walls of a cubic container. The important formula relating pressure (P), volume (V), number of molecules (N), mass of a molecule (m), and the square of the rms speed (v2) of the molecules in a gas is:

PV = Nmv2,

First, we need to determine the volume of the container, which is the cube of one side, so V = (20 cm)3 = (0.2 m)3. Inserting the given values into the equation and solving for P gives us the desired answer. Recall that the rms speed is given, so no temperature calculations are needed.

Therefore, using all given data points:

Volume (V) = (0.2 m)3

Number of molecules (N) = 2.00 × 1023

Mass of one helium molecule (m) = 3.40 × 10-27 kg

Root-mean-square speed (vrms) = 200 m/s

By substituting these values, we can find the pressure exerted by the gas. This represents an application of kinetic theory of gases which assumes the behavior of an ideal gas.

Models can have the same
general appearance as real-ufe
objects.
True or false

Answers

Answer:

true

Explanation:

Answer:

True

Explanation:

Because its true

the reaction A->B is second order in A and the rate constant is 0.039L/mol s. if it took 23 sec for the concentration of A to decrease to 0.30 M, what was the starting concentration of A

Answers

Answer:

0.41 M

Explanation:

A -> B

rate constant (k) = 0.039L/mol s

t = 23

Final concentration, [A] = 0.30M

Initial concentration, [A]o = x

1 / [A] = kt + 1 / [A]o

1 / [A]o = 1 / [A] - kt

1 / [A]o = 1 / 0.30 - 0.039 (23)

1 / [A] = 3.33 - 0.897 = 2.433

[A] = 0.41 M

What is the molarity of the following solutions?a. 19.5 g NaHCO3 in 460.0 ml solution
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution

Answers

Answer:

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

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A 2.00 L container of gas has a pressure of 1.00 atm at 300 K. The temperature of the gas is halved to 150K, and the measured pressure of the same 2.00 Liter sample is 0.420 atm. Which of the following is the best explanation for these observations? A. Pressure is proportional to temperature for a fixed volume of gas. B. The molecules of the gas occupy a significant portion of the volume. C. The molecules of the gas have negligible volume of their own. D. The molecules have significant attractive forces at 150 K. E. The gas is closer to an ideal gas at 150 K.

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Identify the characteristics that describe how human proteins are assembled. Check all that apply.