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Leokris [45]
3 years ago
6

Which information best supports the idea that the universe began with rapid expansion?

Chemistry
2 answers:
koban [17]3 years ago
6 0
Bbbbb x sidisjeiiwkwnsiwowownsxjid i’m so sorry but i need help in this one class i’m trying to get answers to hope you understand
stepan [7]3 years ago
3 0
Answer

I believe it’s A.

Explanation

Because the cosmic microwave background (CMB) is thought to be leftover radiation from the Big Bang, or the time when the universe began
—

The Cosmic Microwave Background radiation, or CMB for short, is a faint glow of light that fills the universe, falling on Earth from every direction with nearly uniform intensity.
You might be interested in
How many atoms in 1kg of platinum<br> a 2.5x10^24<br><br> b 3.1x10^24
Nutka1998 [239]

Answer:

3.1x10^24 it will be in 1 kg of platinum

6 0
3 years ago
What are some natural measures of time
Len [333]

most events like the rising and setting of the Sun were used a natural measurement of time until recently.

Solar time, which is based on the motion of the Sun, is not the only way of measuring time, however. One might keep track of the regular appearance of the full Moon. That event occurs once about every 29.5 solar days. The time between appearances of new moons, then, could be used to define a month.

One also can use the position of the stars for measuring time. The system is the same as that used for the Sun, since the Sun itself is a star. All other stars also rise and set on a regular basis.

Although any one of these systems is a satisfactory method for measuring some unit of time, such as a day or a month, the systems may conflict with each other. It is not possible, for example, to fit 365 solar days into 12 or 13 lunar months exactly. This problem creates the need for leap years

Read more: http://www.scienceclarified.com/Ti-Vi/Time.html#ixzz5e1E705sr

I abbreviated most of it but there is a ton more at this link if you still need more.

7 0
3 years ago
For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
3 years ago
a solution was made by mixing sodium chloride () and water (). given that the mole fraction of water is 0.980 in the solution ob
Mnenie [13.5K]

The mass of sodium chloride used <u>was 1.17 grams</u><u>.</u>

The mole fraction can be calculated by way of dividing the number of moles of 1 factor of an answer via the full variety of moles of all the additives of an answer. it is mentioned that the sum of the mole fraction of all the components inside the solution has to be the same as one.

mass  of NaCl given =  64.9 g

mole  = mass/molar mass

          = 64.9 / 58.5

          =<u> 1.109</u>

a mole fraction of water =  0.980

mole fraction of NaCl = 1 - 0.980

                                   = <u>0.02</u>

1 mole of NaCl =  58.5

mass of NaCl  = 58.5 × 0.02

                       =<u> 1.17 gram</u>

Mole Fraction describes the range of molecules contained within one aspect divided through the whole range of molecules in a given combination. it's miles quite beneficial whilst two reactive-natured components are mixed collectively.

Learn more about mole fraction here:-brainly.com/question/29111190

#SPJ4

4 0
1 year ago
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains
erastova [34]

Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0
3 years ago
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