Please answer any of these
I’ll be grateful and I’m marking brainliest

Answer:

3

Explanation:

sky is high and I am good.

Answer:

Answer:

Explanation:

pretty sure its mixure.

Answer:

0⁰c , 5 mmHg

Explanation:

Because at this temperature we all know water can be in state of both solid and liquid and when pressure is decreased it can be in form of gas too

Answer:

11.92 Liters O₂(g) at STP

Explanation:

For most stoichiometry problems, if one will convert all given data into moles, solve by rxn ratios, then finish by converting to needed dimensions to complete problem.

2Pb(NO₃)₂ => 2PbO + 4NO₂ + O₂

Given 237.5 grams PbO formed in reaction = 237.5 g/223.2 g/mole = 1.064 mole PbO

From equation ratios, moles O₂ formed = 1/2(moles PbO) = 1/2(1.064 moles O₂) = 0.532 mole O₂

0.532 mole O₂ = 0.532 mole x 22.4  liters/mole at STP = 11.91756272 liters (calculator answer) ≅ 11.92 liters O₂ at STP (4 sig.figs. based on given 237.5g PbO)

Teniendo en cuenta que la molaridad es el número de moles de soluto que están disueltos en un determinado volumen, la molaridad de la disolución de HNO₃ es 14.67 [tex]\frac{mol}{L}[/tex].

El porcentaje en masa o en peso de una disolución es el número de gramos de soluto que hay por cada 100 gramos de disolución.

Entonces una pureza de 65% indica que hay 70 g de HNO₃ puro por cada 100 g de HNO₃ impuro.

Por otro lado, la densidad es la relación entre el peso o masa de una sustancia y el volumen que ocupa esa misma sustancia.

Entonces, si la disolución de HNO₃ tiene una densidad de 1.32 g/mL (equivalente a 1320 g/L), donde la cantidad de masa es impura, se puede obtener la cantidad de masa de HNO₃ presente 1 L de la solución mediante el siguiente cálculo:

[tex]\frac{1320 g HNO_{3} impuro}{1 L}x \frac{70 g HNO_{3}puro}{100 g HNO_{3}impuro}= 924 \frac{g HNO_{3}puro}{L}[/tex]

Finalmente, la masa molar de HNO₃, es decir la masa presente en un mol de la sustancia, es 63 [tex]\frac{g}{mol}[/tex].

Entonces, teniendo en cuenta que la molaridad es el número de moles de soluto que están disueltos en un determinado volumen, es posible determinar la molaridad como:

[tex]\frac{924 g HNO_{3}puro}{L}* \frac{1 mol}{63 g}= 14.67 \frac{mol}{L}[/tex]

En resumen, la molaridad de la disolución es 14.67 [tex]\frac{mol}{L}[/tex]

Aprende más sobre la molaridad:

Answer:

Between galaxies, retake the measurement.

Explanation:

Among both galaxies, reiterate the measurements. Instead of utilising a balloon, a scientist might create a model that shows in which everything really is and how cosmos is expanding rather than expanding. By inflating up the atmosphere to 1/2, 3/4, and full size, continue the operation.

Answer:

A

Explanation:

Answer:

fusion is what fuels the burning of the sun,it's the combining of light elements into heavier elements to produce energy.the sun produces a large amount of energy by combining very light elements with heavier elements.

I hope this helps

Answer:

This is true.

Explanation:

R should be 0.0821

Good luck!

Answer:

0.0111moles

Explanation:

as moles = mass/ mr(16+1+1)

so 0.2g / 18

gives you 0.111moles

hope this helps :)

Answer:

the answer is 47.9 and ik because I just had that question

The empirical formula of the compound is O₄Pb₃.

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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Answer:

B.

No products will be formed.

Answer:

The electron from the ground state to occupy a next energy level. In this case,we say that the electron is excited

Answer:

+3

Explanation:

The name of this compound would be nickel (III) oxide. Oxygen typically has a charge of -2. Given there are 3 oxygen atoms, that would equal a charge of -6. The nickel is the cation and must equal the charge of the anion, so must have a charge of +6. Since there are 2 nickel atoms, the charge on each must be +3.

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Answer:

Increasing the temperature

Increasing particle size

Explanation:

Increasing the temperature allows for particles and molecules to move faster (because there is an increase in energy) and allows for more collisions.

∴ Increasing the temperature would be an answer.

Increasing the particle size allows for more surface area and a greater chance for particles to collide with each other.

∴ Increasing particle size would be an answer.

Topic: AP Chemistry

Unit: Kinetics

Answer:

d. 46.5 g

Explanation:

First, you need to start with a properly balanced equation:

N₂H₄(l) + 3O₂(g) -> 2NO₂(g) + 2H₂O(l)

Then, find the moles of water produced:

2H = 2.01568 amu

O = 15.999 amu

________________

18.01468 amu

52.4 g ÷ 18.01468 amu = 2.9087 moles

The mole ratio in our balanced formula between N₂H₄ : H₂O is 1 : 2, so divide moles of H₂O by 2 to get moles of N₂H₄

2.9087 moles ÷ 2 = 1.45435 moles of N₂H₄

Then, calculate the atomic mass of N₂H₄:

2N = 28.0134 amu

4H = 4.03136 amu

________________

32.04476 amu

Finally, calculate the mass in grams of N₂H₄:

1.45435 moles • 32.04476 amu = 46.604 g

The highest energy sublevel of argon will be p.

Argon has an atomic number of 18.

Thus, the electron configuration of argon will be [tex]1s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6[/tex]

The ascending order of energy for the sublevel orbitals is s, p, d, and f.

Therefore, the highest energy sublevel of argon is p.

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The highest energy sublevel in the element argon is p.

The periodic table is an arrangement of atoms according to their order of increasing atomic numbers. Now we know that there are 18 groups in the periodic table and argon belongs to group 18.

The electron configuration of argon is; 1s2 2s2 2p6 3s2 3p6. Thus, the highest energy sublevel in the element argon is p.

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Answer:

The error includes;

a) The wrong denominator for the equivalent fraction of kilograms to grams

b) The use of the scale factor of length rather than the scale factor of volume for the equivalent fraction of cubic centimeters to cubic meters

c) The arrival at the correct 1000 kg/m³ rather than 0.1 g/m³ based on the expression on the left of the equation

Explanation:

The density of water = 1 g/cm³

The given equation is presented as follows;

1 g/cm³ × 1 kg/(1000 kg) × 100 cm³/(1 m³) = 1000 kg/m³

The error in the conversion method are;

a) The conversion, 1 kg/(1,000 kg) has an error, the correct conversion is (1 kg)/1,000 g)

b) The volume conversion error, 100 cm³/(1 m³), the correct volume conversion is (100 cm)³/(1 m³) = 1,000,000 cm³/(1 m³)

c) The right of the equal to sign error; using the left side expression only, the (wrong) answer is 0.1 g/m³

The correct equation is presented as follows;

1 g/cm³ × 1 kg/(1000 g) × 1,000,000 cm³/(1 m³) = 1000 kg/m³

Answer:

C. 1,000 kg should be written as 1,000 g.

Explanation:

Got it right on my test :)

Answer:

A chemical reaction that has a positive ΔH is said to be endothermic , while a chemical reaction that has a negative ΔH is said to be exothermic . What does it mean if the ΔH of a process is positive? It means that the system in which the chemical reaction is occurring is gaining energy.

Answer:

here's the answer to your question

Answer:

Explanation:

C3H8= C3H6 +H2

Answer:

See explanation

Explanation:

Energy of the photon;

E = hc/λ

h= Planks constant

c= speed of light

λ = wavelength

E= 6.6 × 10^-34 × 3 × 10^8/4 x10 -7

E = 4.95 × 10^-19 J

If 1ev=1.602x10-19 J

x = 4.95 × 10^-19 J

x= 3.1 ev

From Einstein's photoelectric equation;

KE = E - Wo

Where;

KE = kinetic energy of ejected photoelectron

E= energy of the photon

Wo= work function of the metal

KE = 3.1 eV - 2.13 eV

KE= 0.97 eV

KE = 0.97 eV × 1.602x10-19 J

KE = 1.55 × 10^-19 J

KE = 1/2 mv^2

1.55 × 10^-19= 1/2 × 9.1 × 10^-31 × v^2

v= √2 × 1.55 × 10^-19/9.1 × 10^-31

v= 5.8 × 10^5 m/s

Answer:

a) clockwise

b) clockwise and

c) anticlockwise

Answer:

1.34 34343 he did

Explanation:

i did the test

Answer:

0.85

Explanation:

C. NaOH

C. Cu (OH)

A. HCl