Please help with whatever this is and go in number order

Answer:  On Earth, liquid water exists on the surface in the form of oceans, lakes and rivers. It also exists below ground as groundwater, in wells and aquifers. Water vapor is most visible as clouds and fog. The frozen part of Earth's hydrosphere is made of ice: glaciers, ice caps and icebergs.

Test tubes heat small amounts of liquids while boiling tube boils liquids

Answer:

B

Explanation:

Answer: 49 joules

Explanation: gravitational potential energy = mgh

m= mass kg,  g= acceleration due to gravity 9.8 m/sec/sec, h= height m

=5*9.8.1 joules = 49 joules

Answer:

Isothermal compressibility is different from isentropic compressibility by temperature instead of entropy.

Explanation:

Isothermal compressibility refers to that type of compressibility where volume change takes place at constant temperature whereas isentropic compressibility refers to that in which volume change takes place at constant entropy. Entropy is the measure of a thermal energy of the system per unit temperature that is unavailable for doing useful work.

Answer:

37.4 mol.

Explanation:

Hello!

In this case, since the Avogadro's number help us to realize that one mole of any substance contains 6.022x10²³ formula units, in this case atoms of zinc, the following dimensional analysis provides the correct answer:

[tex]=2.25x10^{25} atoms*\frac{1mol}{6.022x10^{23}atoms}\\\\= 37.4mol[/tex]

Best regards!

The molecular formula of hydrate : CaCl₂.6 H₂O

So there are 6 molecules of H₂O

Given

54.7g CaCl₂ and 53.64 g H₂O

Required

The number of molecules H₂O

Solution

mol CaCl₂ :

= mass : MW

= 54.7 : 111 g/mol

= 0.493

mol H₂O :

= 53.64 : 18 g/mol

= 2.98

mol ratio H₂O : CaCl₂ :

= 2.98/0.493 : 0.493/0.493

= 6 : 1

Explanation:

first we have to find molar mass of ca(H2c3o2)2

40+(1*2)2+(12*3)2+(16*2)2

40+4+72+64=180g/mole

m=n*Mm

m=1.75mole*180g/mole

m=315g

Answer:

a. ((100 - 35)/100) times 5.5 grams = 3.575 g

Explanation:

Given that:

The sample of carbon and oxygen = 5.5g

where carbon makes 35% of the mass of the substances.

It implies that oxygen will make: (100 - 35)% = 65%

Suppose y be the mass of the oxygen;

Then:

[tex]y = \dfrac{(100-35)}{100} \times 5.5 \ g[/tex]

[tex]y = \dfrac{(65)} {100} \times 5.5 \ g[/tex]

[tex]y = 3.575 \ g[/tex]

The mass of carbon [tex]= \dfrac{35}{100} \times 5.5 \ g[/tex]

= 1.925 g

Answer:

Ideal Gas

The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.

Real Gas

The molecules of real gas occupy space though they are small particles and also have volume.

anation:

The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.

The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.

An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.

On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.

In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.

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Answer:

0 for the elemental form, +2 in its compounds.

The heat released : b. -7.34 x 10² kJ

Given

1.65 mol NaOH

The molar heat of solution of NaOH is -445.1 kJ/mol

Required

Heat released

Solution

ΔH solution = Q : n

ΔH solution = enthalpy of solution(-=exothermic, +=endothermic)

Q = heat released/absorbed

n = moles of solute

Input the value :

Q = ΔH solution x n

Q = -445.1 kj/mol x 1.65 mol

Q = -734.415 kJ

Answer:

Cr

Explanation:

it just is im from Harvard my boy lmk

Answer:

no way

Explanation:

more free points

Answer:

Explanation:

a )

Average reaction rate between 0 and 1500s

Time duration = 1500 s

moles reacted = .184 - .016 = .168 moles

Moles reacted per second = .168 / 1500

= 112 x 10⁻⁶ moles /s

b )

Average reaction rate between 200 and 1200s

Time duration = 1000 s

moles reacted = .129 - .019 = .11 moles

Moles reacted per second = .11 / 1000

= 110 x 10⁻⁶ moles /s

c )

the instantaneous rate of the reaction at t=800s

We shall assume that between 500 s and 1200 s , rate of reaction is uniform

rate between 500 and 1200

Time duration = 700 s

moles consumed = .069 - .019 = .05 moles

Rate of reaction = .05 / 700

= 71 .4 x 10⁻⁶ moles / s

This will also be instantaneous rate of reaction at t = 800 s .

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Answer:

Kp is 0.00177

Explanation:

We state the equilibrium:

CO(g) + Cl₂(g)  ⇆  COCl₂(g)

Initially we have these partial pressures

1.86 atm for CO and 1.27 for chlorine.

During the reaction, x pressure has been converted. As we have 0.823 atm as final pressure in the equilibrium for COCl₂, pressure at equilibrium for CO and chlorine will be:

1.86 - x for CO and 1.27 - x for Cl₂.

And x is the pressure generated for the product, because initially we don't have anything from it. So pressure in equilibrium for the reactants will be:

1.86 - 0.823 = 1.037 atm for CO

1.27 - 0.823 = 0.447 atm for Cl₂

Let's make, expression for Kp:

Partial pressure in eq. for  COCl₂ / P. pressure in eq. for CO . P pressure in eq. for Cl₂

Kp = 0.823 / (1.037 .  0.447) → 0.00177

Answer:

0.229 cm³.

Explanation:

The following data were obtained from the question:

Volume (in in³) = 0.014 in³

Volume (in cm³) =?

1 in = 2.54 cm

Next, we shall determine a conversion scale to convert from in³ to cm³. This can be obtained as follow:

1 in = 2.54 cm

Therefore,

1 in³ = 2.54³ cm³

1 in³ = 16.387 cm³

Finally, we shall convert 0.014 in³ to cm³. This can be obtained as follow:

1 in³ = 16.387 cm³

Therefore,

0.014 in³ = 0.014 in³ × 16.387 cm³ / 1 in³

0.014 in³ = 0.229 cm³

Thus, 0.014 in³ is equivalent to 0.229 cm³.

The percent composition of the compound.

A O-18.18%, N-21.21%, H-60.60%

Given

6.00 grams of oxygen,

7.00 grams of nitrogen,

20.00 grams of hydrogen.

Required

The percent composition

Solution

Total mass :

= mass of O + mass of N + mass of H

= 6 + 7 + 20

= 33 g

% O = 6/33 x 100%= 18.18%

% N = 7/33 x 100%=21.21%

% H = 20/33 x 100% = 60.6 %

Answer:

See explanation

Explanation:

Here we are trying to see how a binary covalent compound is named. A binary covalent compound comprises of only two elements held together by covalent bonds.

The first element retains its normal name whereas the second element has the suffix -ide added to it.

For instance, CO2 is named as carbon dioxide, HBr is named as hydrogen bromide etc.

Answer: it's ide

Explanation:

Got it right in the quiz

Answer:

44

Explanation:

Given that :

Mass of solute = Mass of urea = 16g

Mass of water = 20g

Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g

Percentage Mass = (mass of solute / mass of solution) * 100%

Percentage Mass = (16 / 36) * 100%

Percentage Mass = 0.4444444 x 100%

Percentage Mass = 44.44%

Percentage Mass = 44%

Answer:

Liquid

Explanation:

Answer:

well just like grass that helps animals eat rocks may give an animal a home

Answer:

0.0876L of 0.478M Na₃PO₄ are needed

Explanation:

The reaction of calcium chloride, CaCl₂, with sodium phosphate, Na₃PO₄ is:

3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl

Where 3 moles of calcium chloride react with 2 moles of sodium phosphate to produce 1 mole of calcium phosphate.

To solve this question we need to find moles of CaCl₂ added. Using the reaction we can find the moles of Na₃PO₄ that are needed to react completely and the volume using its concentration:

Moles CaCl₂:

0.225L * (0.279mol / L) = 0.0628moles of CaCl₂

Moles Na₃PO₄:

0.0628moles of CaCl₂ * (2mol Na₃PO₄ / 3 mol CaCl₂) =

0.0419moles of Na₃PO₄

Volume 0.478M Na₃PO₄:

0.0419moles of Na₃PO₄ * (1L / 0.478mol) =

Answer:

[tex]\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

[tex]\Delta H_{rxn} =- m_wC_w\Delta T[/tex]

We plug in the mass of water, temperature change and specific heat to obtain:

[tex]\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ[/tex]

Now, this enthalpy of reaction corresponds to the combustion of propyne:

[tex]C_3H_4+4O_2\rightarrow 3CO_2+2H_2O[/tex]

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

[tex]\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}[/tex]

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

[tex]\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol[/tex]

Now, we solve for the enthalpy of formation of C3H4 as shown below:

[tex]\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}[/tex]

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

[tex]\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]

Best regards!

Answer:So, one mole of water has a mass of 16 +1+1 = 18 grams. So, if one mole has a mass of 18 grams, 25 grams would have a mass of 25 grams/ 18 grams per mole or 1.39 moles

Answer:

The answer would be 1.33

Explanation:

do you need an explanation?

Answer:

See picture below

Explanation:

In this case, is ocurring an ozonolysis reaction with alkines.

Alkines, unlike alkenes, when they undergo an ozonolysis reaction, the product formed is a carboxilic acid. In some cases it may produce CO₂, but that will happen only if the starting molecule has a terminal Hydrogen atom.

In other words the following:

CH₃CH₂C≡CH

In this case, it may produces the CO₂. However,  it's not the case, so it will produce two molecules of carboxylic acid.

You can see the picture below for the final product.

Hope this helps.

Alkynes are subjected to ozonolysis to produce two ketones or acid anhydrides. The acid anhydride undergoes hydrolysis to produce two carboxylic acids if water is present in the process. Elastomer ozonolysis is also referred to as the ozone method.

Ozone (O₃), a reactive allotrope of oxygen, is used in ozonolysis, a process for oxidatively breaking alkenes or alkynes. The procedure enables the substitution of double or triple carbon-carbon bonds with double oxygen bonds. This reaction is frequently used to determine an unknown alkene's structure.

The products of the given reaction can be shown below:

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Answer: Percent recovery is 47.34 %

Explanation:

Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

[tex]{\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100[/tex]

Putting in the values we get:

[tex]{\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%[/tex]

Therefore, the percent recovery is 47.34 %