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1 year ago

6

A group of scientists are studying observable matter in a cluster of galaxies. They conclude that the cluster has only a small p

ercentage of the mass required to keep it from
flying apart. Why does the duster remain intact?

1 answer:

lidiya [134]1 year ago

5 0

Answer:

it is because of Dark Matter

Dark Matter, component of the universe whose presence is discerned from its gravitational attraction rather than its luminosity. Dark matter makes up 30.1 percent of the matter-energy composition of the universe; the rest is dark energy (69.4 percent) and “ordinary” visible matter (0.5 percent).

Dark matter is composed of particles that do not absorb, reflect, or emit light, so they cannot be detected by observing electromagnetic radiation. Dark matter is material that cannot be seen directly.

Explanation:

Hope It helps

Have A Nice Day : )

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How many moles are represented by 3.01 x10^24 oxygen atoms?

asambeis [7]

<h3>Answer:</h3>

5.00 mol O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.<u> </u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.01 × 10²⁴ atoms O₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.01 \cdot 10^{24} \ atoms \ O_2(\frac{1 \ mol \ O_2}{6.022 \cdot 10^{23} \ atoms \ O_2})$
Multiply/Divide: $\displaystyle 4.99834 \ mol \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

4.99834 mol O₂ ≈ 5.00 mol O₂

4 0

2 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

olga55 [171]

Answer: Rate law= $k[A]^1[B]^2$ , order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is $3L^2mol^{-2}s^{-1}$

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

$2=2^x,2=2^1$ , x=1

Thus rate law is $Rate=k[A]^1[B]^2$

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1: $1.2\times 10^{-2}=k[0.10]^1[0.20]^2$

$k=3 L^2mol^{-2}s^{-1}$ .

6 0

2 years ago

A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h

dexar [7]

Answer:

5.758 is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d = $998.2 kg/m^3$

1 kg = 1000 g

$1 m^3=10^6 cm^3$

$998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3$

Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3$

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' = 9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

$v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3$

Density of metal = d'

Volume of metal = v' = $\frac{M'}{d'}$

Difference in volume will give volume of metal ingot.

v' = v - V

$v'=19.73 cm^3-18.08 cm^3=$

$v'=1.65 cm^3$

Since volume cannot be in negative .

Density of the metal =d'

= $d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3$

5 0

2 years ago

How much heat is required to raise the temperature of 50.0g of water 13°C<br>

N76 [4]

Answer:one gram by 1oC

Explanation: you will need to know the value of water's specific heat

3 0

2 years ago

Cho recorded the weather in her town for a week. She wrote down her observations in the table below. Mon Tue Wed Thu Fri Sat Sun

SIZIF [17.4K]

Answer:

If this trend continues, the following week will be cooler, and a large amount of rain will fall.

Explanation:

Patterns and trends can often be found in data sets. During the week that Cho recorded the weather, the temperatures consistently dropped by one to four degrees each day. At the end of the week, the amount of precipitation increased daily.

6 0

3 years ago