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2 years ago

Which statement accurately describes dark matter?

Chemistry

2 answers:

pickupchik [31]2 years ago

6 0

Yes it have to be C

Mandarinka [93]2 years ago

5 0

Answer:

C. It does not emit electromagnetic radiation.

Explanation:

Right now, Dark Matter is only a theory. Scientist proposed this to counter some of the strange phenomenon with matter in space.

Scientists know little about dark matter. Some say it's one of the driving forces of the universe. Currently, scientists have no way of measuring or identifying dark matter.

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What pressure does 3.54 moles of chlorine gas at 376 k exert on the walls of it 51.2 l container? I'm Lazy sooooo

Lilit [14]

Answer:

The answer to your question is P = 2.13 atm

Explanation:

Data

Pressure = ?

number of moles = 3.54

Temperature = 376 °K

Volume = 51.2 L

R = 0.08205 atm L/mol°K

Formula

PV = nRT

- Solve for P

P = nRT / V

- Substitution

P = (3.54)(0.08205)(376) / 51.2

- Simplification

P = 109.21 / 51.2

Result

P = 2.13 atm

6 0

2 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

2 years ago

A chemist adds 180.0 mL of a 1.42M sodium carbonate (Na CO,) solution to a reaction flask. Calculate the millimoles of sodium ca

svet-max [94.6K]

Answer: The millimoles of sodium carbonate the chemist has added to the flask are 256

Explanation:

Molarity is defined as the number of moles dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{milli moles of solute}}{\text{Volume of solution in ml}}$ .....(1)

Molarity of $BaCl_2$ solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in equation 1, we get:

$1.42M=\frac{\text{milli moles of }BaCl_2}{180.0ml}\\\\\text{milli moles of }BaCl_2}={1.42M\times 180.0ml}=256milli mol$

Thus the millimoles of sodium carbonate the chemist has added to the flask are 256.

6 0

2 years ago

If 36.0 g of NaOH (MM = 40.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the

Kobotan [32]

Answer:

Molarity of NaOH = 1.8 M.

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Volume = 500 mL

Molarity of NaOH =?

Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 36 / 40

Mole of NaOH = 0.9 mole

Next, we shall convert 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.9 mole

Volume = 0.5 L

Molarity of NaOH =?

Molarity = mole / Volume

Molarity of NaOH = 0.9 / 0.5

Molarity of NaOH = 1.8 M

8 0

2 years ago

Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution

konstantin123 [22]

Answer:

1.22 mL

Explanation:

Let's consider the following balanced reaction.

2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl

The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol

The molar ratio of AgCl to BaCl₂ is 2:1. The moles of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.

The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:

1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL

8 0

2 years ago

Which statement accurately describes dark matter?￼

Which statement accurately describes dark matter?