What do black holes look like ?
Astronomy class.

What Do Black Holes Look Like ? Astronomy Class.

Answers

Answer 1
black holes cannot be seen, As they consume matter that strays too close, they squeeze it into a superheated disk of glowing gas.
Answer 2
Zzzz was correct give him the crown

Related Questions

3) How many grams of calcium hydroxide are necessary to create 49.5 mL of a 1.55 M solution?

Answers

Answer:

5.68g

Explanation:

What is the mass of 5.55 moles of carbon monoxide

Answers

Answer:

[tex]\boxed {\boxed {\sf 155 \ g\ CO}}[/tex]

Explanation:

To convert from moles to mass, the molar mass must be used.

First, write the chemical formula for carbon monoxide. Since the carbon (C) comes first without a prefix, there is 1 carbon atom. The prefix mono- before oxide means 1, so there is also 1 oxygen (O) atom. The formula is CO.

Next, look up their molar masses on the Periodic Table.

C: 12.011 g/mol O: 15.999 g/mol

Since there is 1 atom of each, the molar masses can be added.

CO: 12.011 g/mol + 15.999 g/mol = 28.01 g/mol

Use this molar mass as a ratio.

[tex]\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

Multiply by the given number of moles:5.55

[tex]5.55 \ mol \ CO *\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

The moles of carbon monoxide cancel.

[tex]5.55 * \frac {28.01 \ g \ CO} {1 }[/tex]

Multiply.

[tex]155.4555 \ g \ CO[/tex]

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, it is the ones place. The 4 in the tenths place tells us to leave the 5.

[tex]155 \ g\ CO[/tex]

5.55 moles of carbon monoxide is about 155 grams.

2 NaOH + H2SO4 → 2 H2O + Na2SO4
How many grams of water are produced from 203.50 grams of H2S04?

Answers

Answer:

[tex]m_{H_2O}=74.8gH_2O[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible to realize there is a 1:2 mole ratio of sulfuric acid to water; thus, given the mass of the former and its molar mass (98.07 g/mol), it is possible to determine the mass of produced water as shown below:

[tex]m_{H_2O}=203.50gH_2SO_4*\frac{1molH_2SO_4}{98.07gH_2SO_4}*\frac{2molH_2O}{1molH_2SO_4} *\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=74.8gH_2O[/tex]

Regards!

By the reaction between
H3PO4 and Ca(OH)2,
Can be formede:

a. Ca(HPo4)2,
b. СaHP4
c. Ca(H2PO4)2
d. Ca2HP02
e.Сa3(PO4)2

the correct answers : b, c, e

the question is:
I am preparing for an admittion test in chemistry, I face this kind of questions which I dont how to understand or if I should only memorize.

I appreciate all your help.
thank you alot.​

Answers

Answer:

ANSWER is E

Explanation:

2H₃PO₄ + 3Ca(OH)₂→ 6H₂O + Ca₃(PO₄)₂

Answer:

First step is to find what the products would be. Since this seems to be a double displacement, you get: H3PO4(aq)+Ca(OH)2(aq)→ H2O(L)+Ca3(PO4)2(s)

You know the products because H has a charge of +1, so PO4 must have a charge of -3, and since OH has a charge of -1, Ca must be +2. Then make each compound have a net charge of 0. H20 is a liquid since aq solutions have liquid water. Ca3(PO4)2 is a solid, use the solubility rules.

Lastly, balance the equation: 2H3PO4(aq)+3Ca(OH)2(aq)→ 6H2O(L)+Ca3(PO4)2(s)

So the salt formed is Ca3(PO4)2

Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you expect
to release upon heating?

Answers

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

[tex] \eta_{KClO_{3}} = \frac{m}{M} [/tex]

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

[tex] \eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles [/tex]                                      

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

[tex] \eta_{O_{2}} = \frac{3}{2}*0.252 moles = 0.378 moles [/tex]

Finally, the mass of O₂ is:

[tex] m = 0.378 moles*32 g/mol = 12.10 g [/tex]

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

What pair of properties do all solids have?

Answers

Answer:

Has definite shape and volume

Solids have high melting point

Explanation:

what is the total of the amount of energy that is reflected back into space​

Answers

moon va half plus 22

1 = 2 Amps, R = 3 Ohms, V = ? Volts​

Answers

Answer:

6 volts

Explanation:

Use Ohm's Law:

V = IR

V = (2 amps)(3 ohms) = 6 volts

6. The drawing shows Samir riding his mountain bike.
i) Draw a circle around the places on the drawing where there
should be a lot of friction.
ii) Explain why there should be a lot of friction in these places.

Answers

Answer:

You Need To Upload The Photo Then I Will Help You :)

Explanation:

The lithosphere ______ part of the earth’s crust.
A. Gaseous
B. Solid
C. Water
D. Liquid

Answers

Answer:

The lithosphere solid part of the earth's crust

What cell tissue helps carry messages? :

options:

muscular
nervous
connective

Answers

Answer:

Nervous

Explanation:

Nervous tissue is composed of two main cell types: neurons and glial cells. Neurons transmit nerve messages.

Most cooking utensils are made up of aluminum because aluminum is _____​

Answers

Explanation:

Cooking utensils, such as pots, pans and menu trays, are often made from aluminium because it is lightweight and conducts heat well, making it energy-efficient for heating and cooling. These properties also make it a preferred material for packaging.

because it is a lightweight material and a good heat conductor

how many grams of P3O9 contain 9.67x10^23 atoms of oxygen?​

Answers

Answer:

49.1 g

Explanation:

First we convert 9.67x10²³ atoms of oxygen into moles, using Avogadro's number:

9.67x10²³ atoms ÷ 6.023x10²³ atoms/mol = 1.60 mol O

Then we calculate how many P₃O₉ moles there would be with 1.60 O moles:

1.60 mol O * [tex]\frac{1molP_3O_9}{9molO}[/tex] = 0.18 mol P₃O₉

Finally we convert 0.18 P₃O₉ moles into grams, using its molar mass:

0.18 mol P₃O₉ * 237 g/mol = 49.1 g

Which carbohydrate strengthens plant cells?
A. glucose
B. glycogen
C. cellulose
D. lactose

Answers

Answer:

a

Explanation:

Answer:

c

Explanation:

edge

How will an increase in the average yearly ocean temperature most
likely affect Earth's weather?

Answers

Answer:

I'd  Say B

Explanation:

Please help asap. Brainliest to correct.

Which formula can be used to calculate the percent yield? (5 points)

Select one:
a. (Theoretical yield ÷ actual yield) x 100

b. (Actual yield ÷ theoretical yield) x 100

c. (Amount of reactants ÷ amount of products) x 100

d. (Amount of products ÷ amount of reactants) x 100

Answers

Answer:

I guess this answer is c

yh

Answer:

b. (Actual yield ÷ theoretical yield) x 100

Explanation:

The percent yield will be a ratio of collected or actual yield to the theoretical yield. Like any percentage, you are determining a part of a whole. You multiply by 100 to convert your decimal answer from the division into a percentage.

How many moles are contained in 325 gram sample of ammonium dichromate (NH4)2Cr2O7

Answers

Answer:

Hyba odpowiedzi c. Hyba

Mole can be defined as the number of molar mass units of the compound in the sample. The moles of ammonium dichromate in 325 grams sample is 1.29 mol.

What is molar mass?

The molar mass can be given as the mass of each atom in the formula unit. The molar mass of ammonium chromate is given as:

[tex]\rm (NH_4)_2Cr_2O_7=2(N)+4\;\times\;2(H)+2(Cr)+7(O)[/tex]

Substituting the mass of each atom to identify the molar mass of ammonium chromate:

[tex]\rm (NH_4)_2Cr_2O_7=2(14)+4\;\times\;2(1)+2(52)+7(16)\\ (NH_4)_2Cr_2O_7=28+8+104+112\\ (NH_4)_2Cr_2O_7=252 \;g/mol[/tex]

The molar mass of ammonium dichromate s 252 g/mol. The moles of the compound in 325 g is given as:

[tex]\rm Moles=\dfrac{mass}{molar\;mass} \\\\Moles=\dfrac{325\;g}{252\;g/mol}\\\\ Moles=1.29\;mol[/tex]

The moles of ammonium dichromate in the sample are 1.29 mol. Thus, option c is correct.

Learn more about moles, here:

https://brainly.com/question/15209553

The total pressure in a container is 37.9 psi. There are two gases in the container. Gas A exerts a pressure of 18.4 psi. What is the pressure of gas B?

Answers

Answer:

19.5

Explanation: using dalton law pt=p1+p2+p3...

the total pressure is 37.9 so to get pressure of gas b subtract pressure of gas a from total pressure.37.9-18.4 gas b equals 19.5

Cells are made of cells true or false

Answers

false

cells are made up of nucleus and cytoplasm and it's contained within the cell membrane

Is the bond length in HCl the same as that in DCl? The wavenumbers of the J = 0  1 rotational transitions for H35Cl and 2H35Cl are 20.8784 and 10.7840 cm–1, respectively. Accurate atomic masses are H = 1.007825 amu, 2H = 2.0140 amu, and 35Cl = 34.96885 amu. Based on this information alone, can you conclude that the bond lengths are the same or different in the two molecules?

Answers

I’m sorry i do not know

how many lone pairs are on the central atom in GeH4?

Answers

It shouldn't have any lone pairs since it is a tetrahedral structure. Ge has 4 valence electrons. Each H has 1 valence electron. Therefore, each H valence electron will pair with each valence electron on Ge.

calculate the wavelength of the first line in the lymen series of hydrogen spectrum​

Answers

since the electron is de-exited from 1(st) exited state (i.e n=2) to ground state (i.e n=1) for first line of Lyman series. Since the atomic number of Hydrogen is 1. λ = 4/3⋅912 A. 1/R = 912 A

A 45 ml sample of nitrogen gas is cooled from 135 degrees celsius to 15 degrees Celsius in a container that can contract or expand at constant pressure. What is the new volume of nitrogen gas?

Answers

Answer:

32 mL

Explanation:

To answer this question we'll use Charles' law, which relates the temperature and volume of a gas at constant pressure:

V₁T₂=V₂T₁

Now we convert the given temperatures to K:

T₁ = 135 °C ⇒ 135 + 273.16 = 408.16 KT₂ = 15 °C ⇒ 15 + 273.16 = 288.16 K

And given that V₁ = 45 mL, we can input the data:

45 mL * 288.16 K = V₂ * 408.16 K

And solve for V₂:

V₂ = 32 mL

A student is given an antacid tablet that weighs 5.4630 g. The tablet is crushed and 4.3620 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 13.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.7 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the stomach acid (in mL) has been neutralized in the 25.00 mL sample that was titrated

Answers

Solution :

It is given that :

Weight of the antacid tablet = 5.4630 g

4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.

25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.

27.7 mL of [tex]$NaOH$[/tex] solution is equivalent to [tex]$25 \ mL$[/tex] of the original stomach acid. Therefore, 13.6 mL of NaOH will take x [tex]$\frac{25\text{ mL of original stomach acid}}{\text{27.7 mL of NaOH}}$[/tex]

                                                             = 12.27 ml of the original stomach acid.

Is this statement true or false?

All gymnosperms lose their leaves in the winter.

Answers

Answer:

they do fall off of themselves

Answer:

True!

Explanation:

They usually fall off in the winter

If a ball rolling down a hill is half way between the top and bottom, how much potential energy does the ball have compared to kinetic energy?

Answers

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength [tex]g[/tex] is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If [tex]m[/tex] denote the mass of this ball, the gravitational potential energy ([tex]\rm GPE[/tex]) of this ball at height [tex]h[/tex] would be [tex]{\rm GPE} = (m \cdot g) \cdot h[/tex], which is proportional to [tex]h\![/tex].

The value of [tex]g[/tex] near the surface of the earth is indeed approximately constant (typically [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex].)

At halfway between the top and bottom of this hill, the height of this ball would be [tex](1/2)[/tex] of its initial value (the value when the ball was at the top of the hill.) Because the [tex]\rm GPE[/tex] of this ball is proportional to its height, at halfway down the hill, the [tex]\rm GPE\![/tex] of this ball would also be [tex](1/2)\![/tex] its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that [tex](1/2)[/tex] of the initial [tex]\rm GPE\![/tex] of this ball was not lost. Rather, these [tex](1/2)\![/tex] of the initial [tex]\rm GPE[/tex] would have been converted to the kinetic energy ([tex]\rm KE[/tex]) of this ball.

Hence, when the ball is halfway down the hill:

[tex]\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}[/tex].

[tex]\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}[/tex].

Therefore:

[tex]\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}[/tex].

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

The dosage for an antibiotic is prescribed at 8.0 mg per kilogram of body weight taken it twice daily for 2 weeks what total mass of antibiotic will be taken by a 180 lb person for the two week period?

Answers

Answer:

18.2784 g

Explanation:

First we convert 180 lb to kg:

1 lb = 0.453 kg180 lb * 0.453 kg / lb = 81.6 kg

Now we calculate the mass of antibiotic taken in one day:

8.0 mg/kg * 81.6 kg = 652.8 mg

Given that it is taken twice daily, in one day the person would take (652.8 * 2) 1305.6 mg. In grams that mass would be:

1305.6 mg / 1000 = 1.3056 g

Finally we multiply those 1.3056 daily grams by the 14 days in two weeks:

(1.3056 g) * 14 = 18.2784 g

The total mass of the antibiotic taken by a 180 lb person for the two-week period would be 18,252.8 mg.

An antibiotic is a type of antimicrobial substance used to treat bacterial infections. It is a medication that either inhibits the growth of bacteria or kills them outright.

Antibiotics work by targeting specific structures or functions in bacterial cells, disrupting their ability to replicate or causing them to die. They are an essential tool in modern medicine for treating various bacterial illnesses and have significantly contributed to improving human health by reducing the impact of infectious diseases.

To calculate the total mass of the antibiotic taken by a 180 lb person for a two-week period,

Convert the person's weight from pounds to kilograms:

Weight in kilograms = 180 lb × 0.453592 kg/lb

= 81.65 kg

Daily dosage = 8.0 mg/kg × 81.65 kg

= 653.2 mg

Total dosage for two weeks = Daily dosage × 2 (twice daily) × 14 (days)

= 18,252.8 mg

Learn more about Antibiotics, here:

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2). H2S and SO2 as follows,(8pt)

H2S + SO2 ------>S + H2O

In a particular experiment, 7.95 g of sulfur is produced by the reaction of 7.5 g of H2S with 12.75 g of SO2.


What is the % yield of sulfur?

Answers

Answer: 75.7 % yield

Explanation:

The balanced chemical equation is:

[tex]2H_2S+SO_2\rightarrow 3S+2H_2O[/tex]  

According to stoichiometry :

68.2 g of [tex]H_2S[/tex] will require = 64 g of [tex]SO_2[/tex]

Thus 7.5 g of [tex]H_2S[/tex] will require = [tex]\frac{64}{68.2}\times 7.5=7.0g[/tex] of [tex]SO_2[/tex]

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 68.2 g of give = 96 g of

Thus 7.5 g of [tex]H_2S[/tex] give =[tex]\frac{96}{68.2}\times 7.5=10.5g[/tex]  of [tex]S[/tex]

% yield=[tex]\frac{\text{Actual yield}}{\text {Theoretical yield}}\times 100=\frac{7.95g}{10.5g}\times 100=75.7\%[/tex]

Thus 75.7 % yield is there.

A 9.56 sample of CuS is found to contain 6.35 g of copper what is the percent by mass of the copy and sulfur in this compound? (show your work)

Answers

Answer:

%Cu = 66.4%

%S = 33.6%

Explanation:

Step 1: Given data

Mass of CuS: 9.56 gMass of Cu: 6.35 g

Step 2: Calculate the mass of S

The mass of CuS is equal to the sum of the masses of Cu and S.

mCuS = mCu + mS

mS = mCuS - mCu

mS = 9.56 g - 6.35 g = 3.21 g

Step 3: Calculate the percent by mass of each element

We will use the following expression.

%Element = mElement / mCompound × 100%

%Cu = mCu / mCuS × 100% = 6.35 g / 9.56 g × 100% = 66.4%

%S = mS / mCuS × 100% = 3.21 g / 9.56 g × 100% = 33.6%


3.
Question
Many chemical warfare reagents readily convert into reactive 3-membered ring intermediates (although
maybe not the Joker's venom from the Batman comics). In turn, these intermediates can damage DNA by
what type of chemical reaction?
Your answer
O a. Elimination
O b. Sn1
OC. SN2
O d. None of the above​

Answers

Answer:

I think it's letter B

Explanation:

Hope it helps correct if I'm wrong

Other Questions
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