How is it possible that we have deserts if the water cycle provides water to earth's surface

Answer:

B. Cell Type

Explanation:

that is the answer i believe

Answer:

option b- cell type

Explanation:

almost absolutely positive

Answer:

Physical and chemical changes can occur almost everywhere, even in our bodies! Food must be broken down into a form that our cells can use. When we eat, our bodies physically break down food into small pieces. Our bodies also chemically break down those small pieces of food into tiny organic molecules.

Explanation:

Answer:

One possible answer could be lemon juice

Explanation:

We had to do a water test, and I was looking up what pH levels were. Then I learned about it in school, so I know a thing or two about it :)

Possible other answers include stomach acid and cola.

Hope this helps!!

Answer:

False

Explanation:

Water vapour exists in gas phase!

Answer:

"The elements are arranged in seven horizontal rows,called periods or series,and 18 vertical columns,called Groups."

Explanation:

Risk*

Answer: 2

Explanation: that’s the answer for the quiz

Answer:

Too high voltage which can't be held by the fuse

Answer:

C. Oxygen

Explanation:

Oxygen has 8 electrons and doesn't react with fluorine noble gases

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Answer:

17.42 g of glucose, C₆H₁₂O₆

Explanation:

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

Mass of water = 300.7 g

Molar solution = 0.322 M

Mass of glucose =.?

Next, we shall convert 300.7 g to L. This can be obtained as follow:

1000 g = 1 L

Therefore,

300.7 g = 300.7 g × 1 L / 1000 g

300.7 g = 0.3007 L

Next, we shall determine the number of mole of glucose in the molar solution. This can be obtained as follow:

Volume of solution = 0.3007 L

Molar solution = 0.322 M

Mole of glucose =?

Molar solution = mole / Volume

0.322 = mole of glucose / 0.3007

Cross multiply

Mole of glucose = 0.322 × 0.3007

Mole of glucose = 0.0968 mole

Finally, we shall determine the mass of glucose, C₆H₁₂O₆, needed to produce prepare the solution. This can be obtained as follow:

Mole of C₆H₁₂O₆ = 0.0968 mole

Molar mass of C₆H₁₂O₆ = (12×6) + (1×12) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ =?

Mole = mass /Molar mass

0.0968 = Mass of C₆H₁₂O₆ / 180

Cross multiply

Mass of C₆H₁₂O₆ = 0.0968 × 180

Mass of C₆H₁₂O₆ = 17.42 g

Thus, 17.42 g of glucose, C₆H₁₂O₆ is needed to prepare the molar solution.

Answer:

Molar mass = 27.88 g/mol

Explanation:

The relationship between how gases effuses is given by Graham's law of effusion. This law is given as;

Rg / Rxe  =   [tex]\sqrt{}[/tex](Mxe / Mg)

Where;

Rg = Rate of effusion of gas g

RXe = Rate of effusion of Xe

Mxe = Molar mass of Xe = 131.29 g/mol

Mg = Molar mass of gas g

From the question;

Rg : Rxe = 2.17 : 1

Rg / Rxe = 2.17 / 1 = 2.17

2.17 = [tex]\sqrt{}[/tex]131.29 / Mg

Squaring both sides

4.7089 = 131.29 / Mg

Mg = 131.29 / 4.7089

Mg = 27.88 g/mol

Answer: 1 mole of strontium, 2 moles of hydrogen, 2 moles of carbon, 6 moles of O

Explanation: i think i’m not sure

A mole corresponds to the mass of a substance that contains particles of the substance.

The concept of moles are used in various ways:-

According to the question, the formula required is:-

[tex]n=\frac{Number\ of\ molecules}{na}[/tex]

The Na is defined as the Avogadro constant is the proportionality factor that relates the number of constituent particles in a sample with the amount of substance in that sample.

According to the question, the compound is [tex]Sr(HCO_{3})_{2}[/tex].

Hence the correct option is 2, that is 1 mole of strontium, 2 moles of hydrogen, 2 moles of carbon, 6 moles of oxygen.

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Answer:

plants

Explanation:

because it produces ( i am a third grade student and don't know if it is right or not)

Answer:

a. 0.528 M .

Explanation:

Hello!

In this case, since the given by-mass percent can be written as:

[tex]\frac{10gCdSO_4}{100g\ sol}[/tex]

By using the density and molar mass of the solute, cadmium sulfate, we can compute the molarity, by also making sure we convert from mL to L of solution:

[tex]M=\frac{10gCdSO_4}{100g\ sol}*\frac{1molCdSO_4}{208.46gCdSO_4} *\frac{1.10g\ sol}{1mL\ sol}*\frac{1000mL}{1L} \\\\ M=0.528M[/tex]

Thereby, the answer is a. 0.528 M .

Best regards.

The molarity of the solution of 10% by mass cadmium sulfate [tex](CdSO_4)[/tex] is approximately 0.479 M. The correct option is C.

To calculate molarity we need to find out how many moles of CdSO4 are present in the solution.

Given:

We need to calculate the mass of [tex]CdSO_4[/tex]:

Mass of [tex]CdSO_4[/tex] = (10% / 100%) * Total mass of the solution

Mass of [tex]CdSO_4[/tex] = (10 / 100) * 1000 g (since the volume is 1 L, and the density is 1.10 g/mL)

Mass of [tex]CdSO_4[/tex] =  100 g

So, the number of moles of CdSO4:

Number of moles of [tex]CdSO_4[/tex] = Mass of CdSO4 / Molar mass of CdSO4

Number of moles of [tex]CdSO_4[/tex] = 100 g / 208.46 g/mol

Number of moles of [tex]CdSO_4[/tex] ≈ 0.479 moles

Then, we calculate the molarity of the solution:

Molarity = Number of moles of CdSO4 / Volume of the solution (in liters)

Molarity = 0.479 moles / 1 L

Molarity ≈ 0.479 M

Hence, the molarity of the solution of 10% by mass cadmium sulfate [tex](CdSO_4)[/tex] is approximately 0.479 M. The correct option is C.

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Answer:

Mass = 99.2 g

Explanation:

Given data:

Volume of oxygen = 15 L

Mass of oxygen = ?

Pressure of oxygen = 75 psi ( 75/14.696 = 5.10 atm)

Temperature of oxygen = 28°C (28+273 = 301 K)

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

n = PV/RT

n = 5.10 atm× 15 L / 0.0821 atm.L/ mol.K   × 301 K

n = 76.5 atm.L/24.71 atm.L/ mol

n = 3.1 mol

mass of oxygen:

Mass = number of moles × molar mass

Mass = 3.1 mol × 32 g/mol

Mass = 99.2 g

Answer:

i just took it the answers is C

Explanation:

Answer:

Option D: Cations

Explanation:

Have a great day.

Answer:

yeye

Explanation:

eq

Answer:

[tex]h=12.9cm[/tex]

Explanation:

Hello!

In this case, since we can consider the beaker until the 100-mL mark as a cylinder, we can use the following equation to relate its diameter, vertical distance or height and volume:

[tex]V=\pi h\frac{d^2}{4}[/tex]

Thus, since we know the diameter, volume (which is equivalent to 600 cm³) and π, we can plug in to obtain:

[tex]600cm^3=\pi *h*\frac{(77.0mm)^2}{4}[/tex]

It means it is necessary to take the mm to cm and solve for h:

[tex]h=\frac{600cm^3}{\pi*\frac{(7.70cm)^2}{4}} \\\\h=12.9cm[/tex]

Best regards!

The distance between each 100 mL mark is 2.15 cm.

The volume of a cylinder is obtained using the formula;

V = πr^2h

Now, we have the following information;

Volume of the cylinder = 600. mL or 600 cm^3

Diameter of the cylinder = 77 mm or 7.7 cm

Radius of the cylinder = 7.7/2 = 3.85 cm

Height of the cylinder  = h

Hence;

600 = 3.142 × ( 3.85 )^2 × h

h = 600/3.142 × ( 3.85 )^2

h = 12.88 cm

There are six 100 mL marks on the beaker, the distance between each 100 mL mark =  12.88 cm/6 = 2.15 cm

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Answer:

1-Ecologists call this one-way flow of matter and energy from a producer to a primary, then to secondary, and perhaps to a tertiary consumer, a food chain. The food web shows all the organisms at each feeding level or trophic level.

2-Energy flows through an ecosystem in one direction: from producer to primary consumer to secondary consumer to higher consumers. A biomass pyramid provides a picture of the feeding and energy relationships within a food web and the direction of the flow of energy.

Answer:

food

Explanation:

Answer:

C3H8 + 5O2 ‐‐‐‐‐‐‐‐> 4H2O + 3CO2

Answer:

Hence the name of the compound is vanadium (III) sulfide

The chemical symbol for the vanadium(III) sulfide compound is [tex]V_2S_3[/tex]. In this chemical, vanadium (V) is found in +3 oxidation state and sulfur (S) is found in the form of sulphide ions.

The compound contains two vanadium ions, as indicated by the numerical prefix "di-" before the element symbol V. An inorganic compound having a crystal structure that produces a solid is called vanadium(III) sulfide.

It is composed of sulfide anions and vanadium cations arranged in a certain way. Vanadium(III) sulfide has many uses in various industries. It can be used as a catalyst in chemical reactions, as part of some types of batteries, and as a raw material for batteries to produce ceramics and materials with specific electrical and magnetic properties.

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Answer:

C. CaCO3

D. CO2

E. NaCl

. H2O

Answer:B because all of the atoms exploded AND GRAVITY WAS CREATED AND forced molecules together (God caused it btw)

Explanation:The BIBLE

Answer:

21. Al (Aluminium)

+3 (most common), -1, -2, +1, +2

22. Bi (Bismuth)

+3 (most common), -3, -2, -1, +1, +2, +4, +5

23. Ni (Nickel)

+2 (most common), -2, -1, 0,+1, +3, +4

24. Xe (Xenon)

0 (most common), +1, +2, +4, +6, +8

25. Pb (Lead)

+2 (most common), +4 (most common), -4, -2, -1, +1, +3

Explanation:

Now we proceed to present the answers for each element:

21. Al (Aluminium)

+3 (most common), -1, -2, +1, +2

22. Bi (Bismuth)

+3 (most common), -3, -2, -1, +1, +2, +4, +5

23. Ni (Nickel)

+2 (most common), -2, -1, 0,+1, +3, +4

24. Xe (Xenon)

0 (most common), +1, +2, +4, +6, +8

25. Pb (Lead)

+2 (most common), +4 (most common), -4, -2, -1, +1, +3

Answer:

50%

Explanation:

Let assume that there are:

"a" moles of CH4 & "b" moles of C2H2

Then by applying the ideal gas equation:

PV = nRT

[tex]16.8 \times 10^3 \times V = (a+b)RT[/tex]

Make (a+b) the subject of the formula:

[tex](a+b) = \dfrac{16800 \ V }{RT} mol[/tex] --- (1)

Since 1 mole of CH₄ yields 1 mol of CO₂ & 1 mol of C₂H₂ yields 2 moles of CO₂

Then;

the total moles of CO₂ = (a +2b)

Now:

[tex]25.3 \times 10^3 \times V = (a+ 2b)RT[/tex]

[tex](a + 2b) = \dfrac{25200 \ V}{RT \ mol}[/tex]    ---- (2)

∴

By solving the above equations

[tex]a = \dfrac{8400 \ V}{RT} \\ \\ b = \dfrac{8400 \ V}{RT}[/tex]

Hence, the estimate of the percentage of methane in the original mixture is:

[tex]= 100 \times \dfrac{a}{(a+b)}[/tex]

[tex]= 100 \times \dfrac{8400\dfrac{ \ V}{RT} }{8400\dfrac{ \ V}{RT} + 8400\dfrac{ \ V}{RT}}[/tex]

= 50%

Answer:

B) 2

Explanation:

There are two significant figures in given measurement,

2, 0

Significant figures:

1= The given measurement have four significant figures 1534.

2= All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

3= Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

4= Zero between the non zero digits are consider significant like 107 consist of three significant figures.

5= The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

Answer:

78.4g

Explanation:

Given parameters:

Mass of copper (II) chloride  = 90g

Mass of sodium nitrate  = 120g

Unknown:

Mass of sodium chloride that can be formed  = ?

Solution:

The balanced chemical reaction is:

            CuCl₂   +   2NaNO₃ →   Cu(NO₃)₂   +   2NaCl

Let us find the limiting reactant. This reactant will determine the extent of the reaction of the amount of product that will be formed.

First, convert the masses to number of moles;

      Number of moles  = [tex]\frac{mass}{molar mass}[/tex]

Molar mass of   CuCl₂  = 63.5 + 2(35.5) = 134.5g/mol

Molar mass of NaNO₃   = 23 + 14 + 3(16) = 85g/mol

  Number of moles of CuCl₂ = [tex]\frac{90}{134.5}[/tex]  = 0.67mole

  Number of moles of NaNO₃  = [tex]\frac{120}{85}[/tex]  = 1.41mole

From the balanced reaction equation:

      1 mole of CuCl₂   would react with 2 moles of NaNO₃

     0.67mole of CuCl₂   would require   0.67 x 2  = 1.34mole of NaNO₃

So, CuCl₂ is the limiting reactant

     1 mole of CuCl₂ will produce 2 mole of NaCl

    0.67mole of CuCl₂ will therefore yield 0.67 x 2 = 1.34mole of NaCl

Mass of NaCl = number of moles x molar mass

  Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl  = 58.5 x 1.34  = 78.4g