Which statement best describes a waxing gibbous moon?

The moon has moved past the new moon and is moving toward the first quarter.

The moon has moved past the first quarter and is moving toward the full moon.

The moon has moved past the third quarter and is moving toward the new moon.

The moon has moved past the full moon and is moving toward the third quarter.

Answer:

mass= 150 g

volume= 550 cm3

density= mass/ volume

= 150/550

= 0.27 g/cm3

Answer: Either [tex]C_5H_6O[/tex]  or [tex]C_{10}H_{12}O_2[/tex]

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

The empirical formula is [tex]C_5H_6O[/tex]

Now we have to calculate the molecular formula:

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}[/tex]

n is a whole number

The molecular formula could be=[tex]1\times C_5H_6O=C_5H_6O[/tex]

Thus molecular formula could be = [tex]2\times C_5H_6O=C_{10}H_{12}O_2[/tex]

Answer:

7. 6 atoms po

Explanation:

sana makatulong po

The question is incomplete, the complete question is;

3CsBr + Al(OH)3 → AlBr3 + 3CsOH

If 25 g of AlBr3  was produced when 125 mL of a CsBr solution was added, what was the molarity of the CsBr solution?

Answer:

2.25 M

Explanation:

Number of moles of AlBr3   produced = 25g/266.69 g/mol = 0.0937 moles of AlBr3  .

Now, from the reaction equation;

3 moles of CsBr yields 1 mole of AlBr3

Therefore, x  moles of CsBr  will yield 0.0937 moles of AlBr3  

x = 3 * 0.0937 = 0.2811 moles

If 125 mL of a CsBr solution was added

And;

n =CV

where;

n = number of moles

C = concentration of solution

V = volume of solution

C = n/V

C = 0.2811 moles * 1000/125

C= 2.25 M

Answer:

1.16 cubic centimeters

Explanation:

Answer:

C 1,868 years

Explanation:

I just took the test - I tried doing the math and was close but couldn't get it exact; so I guessed and I was correct :)

Answer:

1,868 years

Explanation:

Answer:

V = 5.6 L

Explanation:

Given data:

Number of moles of C₄H₁₀ = 0.25 mol

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Volume of C₄H₁₀ = ?

Solution:

Formula:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm × V = 0.25 mol ×  0.0821 atm.L /mol.K × 273.15 K

V = 5.6 atm.L/ 1 atm

V = 5.6 L

Answer:

(1). The vapor pressure is 91 mmHg at 20°C.

(2). No, benzene will not boil at sea level.

Explanation:

Benzene, C6H6 is an aromatic, liquid compound with with molar mass of 78.11 g/mol and Melting point of 5.5 °C. One of the importance or the uses of benzene is in the making of fibres and plastics.

The vapour pressure of benzene can be gotten from the table showing the vapor pressure of different liquids.

Boiling point can simply be defined as the point or the temperature in which the vapor pressure is the same with the atmospheric pressure.

The atmospheric pressure is 760mmHg, while the vapor pressure at sea level is at the temperature of 15°C which is equal to 71 mmHg( from the table showing the vapor pressure of different liquids).

71 mmHg is not equal to 760 mmHg, thus, at sea level Benzene will not boil.

A benzene is chemical that is color less or light to the eyes and is liquid at the rom temperature. It has a sweaty odor and is highly flammable.  

Thus the vapor pressure is 91 mmHg at 20°C, and No, benzene will not boil at the sea level.

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Answer:

Ba(NO₃)₂(aq) + Na₂SO₄(aq) ⇒ BaSO₄(s) + 2 NaNO₃(aq)

Explanation:

Let's consider the unbalanced double displacement equation for the reaction of an aqueous solution of barium nitrate and an aqueous solution of sodium sulfate.

Ba(NO₃)₂(aq) + Na₂SO₄(aq) ⇒ BaSO₄(s) + NaNO₃(aq)

To get the balanced equation, we will multiply NaNO₃(aq) by 2.

Ba(NO₃)₂(aq) + Na₂SO₄(aq) ⇒ BaSO₄(s) + 2 NaNO₃(aq)

Answer:

.133

Explanation:

Dissolution of solute lowers the vapor pressure of solvent .

Relative lowering by non -volatile solute is equal to mole fraction of solute dissolved .

Vapor pressure of pure water is 2.33 k Pa

Lowering of vapor pressure = 2.33 - 2.02 = .31 k Pa .

relative lowering = .31 / 2.33 = .133

relative lowering will be equal to mole fraction of non volatile solute like NaCl

so mole fraction of NaCl dissolved = .133

Answer:

Explanation:

λ  =  [tex]10^8[/tex]   meters

Explanation:

f : frequency

c :  velocity of light

λ  : wavelength

f  =  c/λ

λ  =  c /f

c  =  2.99  ⋅ [tex]10^8[/tex] m/ s

λ  =  [tex]10^8[/tex] meters

Answer:

C. 43 kJ

Explanation:

Bond energy (bond enthalpy (H) or bond strength) is the amount of energy required to break a bond. The bond energy is used to determine the strength of a chemical bond.

The bond energy is the difference between the energy of the reactants and the energy of the products. The bond energy is given by:

[tex]\Delta H=\Sigma Bond\ energy\ reactants-\Sigma Bond\ energy\ products\\\\Where \ \Delta H=bond\ energy\ or\ energy\ released\\\\Given\ that\ \Delta H=25\ kJ,\Sigma Bond\ energy\ reactants=68\ kJ, Hence:\\\\\Delta H=\Sigma Bond\ energy\ reactants-\Sigma Bond\ energy\ products\\\\Substituting:\\\\25\ kJ=68\ kJ- \Sigma Bond\ energy\ products\\\\ \Sigma Bond\ energy\ products=68\ kJ-25\ kJ\\\\ \Sigma Bond\ energy\ products=43\ kJ\\[/tex]

Answer: is simply C.

Answer:

The frequency of the light is 6.62 * 10¹⁴ Hz

Explanation:

The frequency of a wave is defined as the number of complete oscillations made by the wave per second. Its unit is Hertz. The frequency of light as well as the velocity of the light and wavelength of the light are all related.

The velocity of light is the distance the light wave travels per second.

Wavelength is defined as the distance between successive crests and troughs of the wave. It is inversely related to the frequency of the wave.

The velocity of light, c is related to frequency, f, and wavelength, λ, by the formula, c = λf

where the velocity of light, c is a constant = 3 * 10⁸ m/s

λ is wavelength of light = 453 nm = 4.53 * 10⁻⁷ m

Frequency, f = ?

Making f subject of formula in the equation above, f = c/λ

f = 3 * 10⁸ m/s / 4.53 * 10⁻⁷ m

f = 6.62 * 10¹⁴ Hz

Therefore, the frequency of the light is 6.62 * 10¹⁴ Hz

Answer:

1s2 2s2 2p6 3s2 3p6

Explanation:

Phosphorus is a non metal. It gains electrons to form ions. The most likely ion formed is the phosphide, where it gains three extra electrons.

Total number of electrons in Phosphide = 15 + 3 =18

The electronic configuration is given as;

1s2 2s2 2p6 3s2 3p6

Electronic Configuration can be defined as the electron distribution of an atom or molecule in an atomic or molecular orbit. The atomic orbit consists of the s, p and d orbitals.

The Atomic number of Phosphorus is 15.

Thus, the Electronic configuration of Phosphorus is 1s² 2s² 2p6 3s² 3p³.

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5.

Solid Phase (Ice)

Unlike most other substances, water's solid form (ice) is less dense than its liquid form, as a result of the nature of its hexagonal packing within its crystalline structure. ... However, a block of ice floats in liquid water because ice is less dense than liquid water.

Unlike most other substances, water's solid form (ice) is less dense than its liquid form, as a result of the nature of its hexagonal packing within its crystalline structure. This lattice contains more space than when the molecules are in the liquid state. ... Liquid water is most dense at 4 degrees Celsius.

6.Evaporation is a phase transition from the liquid phase to the gas phase that occurs at temperatures below the boiling point at a given pressure.

Answer:  [tex]\Delta H[/tex] for the reaction is 87935 J

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]

[tex]moles of HCl[/tex]=[tex]Molarity\times {\text {Volume in L}}=1.35\times 0.01=0.135moles[/tex]

[tex]moles of NaOH[/tex]= [tex]Molarity\times {\text {Volume in L}}=1.76\times 0.01=0.176moles[/tex]

As 1 mole of HCl neutralizes 1 mole of NaOH

0.135 moles of HCl will neutralize = 0.135 mole of NaOH

Thus HCl is the limiting reagent.

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

Q = Heat absorbed by water = ?

m = mass of water = 200 g    (volume of water=200ml, density of wtaer = 1 g/ml)

c = specific heat capacity = [tex]4.18 J/g^0C[/tex]

Initial temperature of water = = 24.6°C

Final temperature of water=  = 38.8 °C

[tex]Q=200\times 4.18\times (38.8-24.6)[/tex]

[tex]Q=11871.2J[/tex]

Thus heat released by 0.135 moles is = -11871.2 J

heat released by 1 mole is =[tex]\frac{-11871.2 J}{0.135}\times 1=-87935J[/tex]

The heat of reaction, ΔH is 87935 J

Equation of reaction between HCl and NaOH is given below:

Molarity = number of mole/volume

Number of moles = molarity * volume

Number of moles of  HCl = 1.35 M * 0.1 L = 0.135 moles

Moles of NaOH = 1.76 M * 0.1 L = 0.176 moles

1 mole of HCl neutralizes 1 mole of NaOH

Calculating the quantity of heat evolved:

where m is mass, c is specific heat capacity, ΔT is temperature change

Heat absorbed by water, Q = mcΔT

mass of water = density * volume

density of water = 1 g/mL

volume of water = 200 mL

mass of water = 1 g/mL * 200 mL

mass of water = 200 g

specific heat capacity of water, c = 4.18 J/g°C

Initial temperature of water = = 24.6°C

Final temperature of water=  = 38.8 °C

ΔT = 38.8 - 24.6 = 13.7°C

Q = 200 g * 4.18 J/g°C * 14.2°c

Q =  11871.2 J

Therefore, heat released by 0.135 moles is = -11871.2 J

Heat released by 1 mole = 11871.2 * 1/0.135

Heat released by 1 mole = 87935 J

Therefore, the heat of reaction, ΔH is 87935 J

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Answer:

5.0 × 10²⁴ molecules

Explanation:

Step 1: Write the balanced double displacement reaction

2 NaOH + CuSO₄ ⇒ Na₂SO₄ + Cu(OH)₂

Step 2: Calculate the moles corresponding to 5.0 × 10²⁴ molecules of Na₂SO₄

We will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

5.0 × 10²⁴ molecule × 1 mol/6.02 × 10²³ molecule = 8.3 mol

Step 3: Calculate the moles of CuSO₄ required to produce 8.3 moles of Na₂SO₄

The molar ratio of CuSO₄ to Na₂SO₄ is 1:1. The moles of CuSO₄ required are 1/1 × 8.3 mol = 8.3 mol.

Step 4: Calculate the molecules corresponding to 8.3 moles of CuSO₄

We will use Avogadro's number.

8.3 mol × 6.02 × 10²³ molecule/1 mol = 5.0 × 10²⁴ molecule

The mass of 63 ml sample : 79.38 g

Given

20 ml and 25.2 g of glycerol

Required

The mass of 63 ml sample

Solution

Density is the ratio of mass per unit volume

Density formula:

[tex]\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}[/tex]

Density of glycerol :

= m : V

= 25.2 g : 20 ml

= 1.26 g/ml

Mass of 63 ml sample :

= density x volume

= 1.26 g/ml x 63 ml

= 79.38 g

Answer:

n-Pentane

Explanation:

All molecules possess dispersion forces. It is possible in non-polar molecules that a temporary distortion in the electron distribution of one molecule can affect the electron distribution in a neighboring molecule. This is known as  induced dipole, induced dipole forces.

However, this  induced dipole, induced dipole forces depends on the shape of the molecule. Molecules of n-pentane are capable of better interaction via  induced dipole, induced dipole forces since the molecules interact over the entire length of the molecule rather than neopentane which is more compact and nearly spherical.

Answer:

There will be an increased span of the melting point of acetanilide

Explanation:

if benzoic acid and acetanilide are mixed in equal amounts, there will be an increased span of the melting point of acetanilide than what is known already. This is because the two compounds have mixed together to become a mixture.

Answer:

Au

Explanation:

For the density of a face-centered cubic:

[tex]Density = \dfrac{4 \times M_w}{N_A \times a^3}[/tex]

where

[tex]M_w[/tex] = molar mass of the compound

[tex]N_A=[/tex] avogadro's constant

[tex]a^3 =[/tex] the volume of a unit cell

Given that:

Density [tex](\rho)[/tex] = 19.30 g/cm³

a = 0.408 nm

a = [tex]0.408 \times 10^{-9} \times 10^{2} \ cm[/tex]

a = [tex]4.08 \times 10^ {-8} \ cm[/tex]

∴

[tex]19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}[/tex]

[tex]M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}[/tex]

[tex]M_w=197.37 \ g/mol[/tex]

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

Answer:

Gelada baboons plays a significant role.

Explanation:

The role of gelada baboons in their ecosystem is very important because they aerate the soil for plants which is necessary for good plant growth. These gelada baboons also helps in controlling the population of predator in their ecosystem which is very essential for the stability and equilibrium of the ecosystem so gelada baboons has a good effect on both plants and animals in their ecosystem.

Answer: Primary Consumer

Explanation: Gelada Baboons are herbivores, :D

Answer:

Strontium is a soluble earth metal with the nuclear number 38. Phosphate is a polyatomic particle containing phosphorus and oxygen molecules. Strontium loses electrons to turn out to be emphatically charged, and phosphate is an adversely charged particle.

Explanation:

When strontium phosphide comes into contact with water, it transforms into a solid that emits poisonous and combustible fumes. It is a highly reactive substance utilized in the production of chemically reactive devices as well as a laboratory reagent. Strontium phosphide is ionic.

Two elements combine to form an ionic compound. Sr is the metal while P is the non-metal in the ionic combination. After transferring its two valence electrons to P to create the phosphide ion, which has a net negative charge of -3, strontium becomes a cation.

Ions are charged particles that form when an atom (or group of atoms) receives or loses an electron, and ionic compounds are composed of these charged particles.

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Answer:

d. selenium (Se) I’m a non metal

Explanation:

What we know about the element in view is that it has:

              4 orbitals

              6 valence electrons

The orbitals are the energy levels in the element.

Valence electrons are the outermost shell electrons.

The orbitals tells us the periods in which an element belongs. Since we have 4 orbitals, this suggests that our element belongs to the 4th period.

The number of valence electrons tells us the group in which the element could be found. Since we have 6 valence electrons, the group number is 6.

Therefore, the element is period 4 and group 6 is selenium. Selenium is a non-metal.

Answer:

When 150.0g of the solution are evaporated, 7.50g of KCl must be collected

Explanation:

A solution 5% of potassium chloride means that there are 5g of KCl per 100g of solution.

When all the solution is evaporated, the solute KCl will be collected as a solid.

In 150g of solution there are:

150g * (5.00gKCl / 100g solution) = 7.50g of KCl are present in 150g of solution.

That means:

Answer:

7.07

Explanation:

HA = weak acid = 0.053

A+ = conjugate base = 0.045

Ka = 7.2x10^-8

Ka = [H+][A-]/HA

7 2x10^-8 = [H+][0.045]/0.053

[H+] = 7.2x10^-8 x 0.053/0.045

= 8.48x10^-8

PH = -log[H+]

= -log[8.48x10^-8]

PH = -[login.48 + log10^-8]

PH = -0.928 - (-8)log10

= 7.07

Answer:

×

Explanation:

FALSE BECAUSE NOT OCEAN IT IS ALL

Answer:

Anwser A

Explanation:

Sand dunes are created by wind deposition.

The accumulation of sand that produce the sand dunes is called wind deposition. Therefore, option (A) is correct.

Sand dunes are formed when wind deposits the sand on top of each other. It stays until a small mound starts to form. As the first mound forms, sand builds up on the windward side more and more until the edge of the dune breaks due to its own weight.

When the angle of the dune becomes too big to support the weight,  the collapse of sand dune begins. This is called the angle of repose usually around 30°-34° also depends upon wind speed, grain size, and roundedness of the individual grains.

As these dunes collapse, the sand starts to move in the same direction as the wind. The sand piles up on multiple sides of the dune by changing wind direction and is unable to get the angle of repose such dune never collapses.

Therefore, the sand dunes produce the accumulation of sand through a process called wind deposition.

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