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Why must many of the telescopes that detect invisible wave, such as infrared, microwaves, and gamma rays, be placed in space?



a Most radiation reaching Earth from space is blocked by the atmosphere. Therefore, some telescopes must be placed above the atmosphere.


b There are not enough places on land to build observatories that hold large telescopes.


c All telescopes must be placed in space. There are not telescopes on land.


d The waves in the electromagnetic spectrum are all too small to be detected by telescopes on land. Therefore, in order to see any EM radiation, telescopes must be placed in space.

Answer:

they do fall off of themselves

Answer:

True!

Explanation:

They usually fall off in the winter

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!

Answer:

Has definite shape and volume

Solids have high melting point

Explanation:

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.

Answer:

%Cu = 66.4%

%S = 33.6%

Explanation:

Step 1: Given data

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%

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.

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

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.

Answer:

I'd  Say B

Explanation:

Answer:

I think it's letter B

Explanation:

Hope it helps correct if I'm wrong

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.

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

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.

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

Answer:

6.88%

Explanation:

Step 1: Given data

Step 2: Calculate the mass of the solution

The mass of the solution is equal to the sum of the masses of the solute and the solvent.

mSolution = mSolute + mSolvent

mSolution = 27.7 g + 375 g = 402.7 g

Step 3: Calculate the percentage by mass of CaCl₂

We will use the following expression.

%m/m = mass CaCl₂/mass Solution × 100%

%m/m = 27.7 g/402.7 g × 100% = 6.88%

false

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

Answer:

true

Explanation:

the clear lens over the eye is called the cornea

Answer:

Nervous

Explanation:

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

Answer:

6 volts

Explanation:

Use Ohm's Law:

V = IR

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

Answer:

a

Explanation:

Answer:

c

Explanation:

edge

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:

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

And solve for V₂:

Answer:

a. 1.21M

b. 0.119M

c. 0.00496M

Explanation:

Molarity, M, is an unit of concentration defined as the ratio between moles of solute and liters of solution:

a. 4.35 mol LiCl / 3.60L = 1.21M

b. 29.43gC6H12O6 * (1mol / 180.16g) = 0.1634moles / 1.37L = 0.119M

Molar mass C6H12O6: 180.16g/mol

c. 34.5mg NaCl = 0.0345g * (1mol / 58.44g) = 5.9x10⁻⁴moles / 0.1191L = 0.00496M

Answer:

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

Explanation:

Answer:

The lithosphere solid part of the earth's crust

Answer:

5.68g

Explanation:

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

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!

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.

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.