how would you know if the material is harmful or useful to you ?​

The term "molar volume" (Vm) of an ideal gas refers to this amount. (STP) as 293 kelvin (K) of temperature and 1 atmosphere of pressure (P = 1 atm, which is also equal to 760 torr). Any ideal gas has a 22.4 L molar volume at STP.

The volume occupied through one mole of the a chemical element or chemical compound at a standard temperature and pressure (STP) is known as the molar volume (Vm). By dividing the mass density () by the molar mass (M), it can be computed.

One mole of any gas has the same volume when it is present at the same temperature and pressure. The volume that one mole of any gas occupies at standard pressure and temperature is known as the molar volume. 24 dm3 is equivalent to the molar volume (24,000 cm 3). In cases where it is necessary, this volume is provided.

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Answer: we could tell Ca(NO3)2 and Zn(NO3)2 apart from their reactions with KOH and Na2C2O4, based on the color of the precipitate formed and the speed which each precipitate is formed. Also, a reaction with NaOH could be used and the solubility of the formed precipitate in excess NaOH could also be used to differentiate Ca(NO3)2 and Zn(NO3)2.

Explanation:

The question requires us to determine how to tell apart solutions of Ca(NO3)2 and Zn(NO3)2, based on the table of informations provided.

(General topic: Solutions)

According to the table provided by the question, both Ca(NO3)2 and Zn(NO3)2 do not react with KI and Na2SO4 (we could also say that the reaction of Ca(NO3)2 and Zn(NO3)2 with KI or Na2SO4 do not produce any changes that we can identify). On the other side, the reaction of these nitrate salts with KOH and Na2C2O4 produce precipitates with similar characteristics (all of them are white precipitate, with small differences among them).

We could differentiate Ca(NO3)2 from Zn(NO3)2 based on these characteristics:

1) The color of precipitates formed when reacting the nitrate salts with KOH is slightly different (faint white for calcium nitrate and milky white for zinc nitrate). Also, the reaction between Ca(NO3)2 and KOH is slower, then the precipitate formed when reacting Zn(NO3)2 with KOH would form faster;

2) Similarly, the color of the precipitates formed when reacting the nitrate salts with Na2C2O4 are sloghtly different and the reaction of Zn(NO3)2 with Na2C2O4 is slower, therefore the precipitate formed in this reaction would be produced slower (or wouldn't be observed due to the slow reaction);

3) Additionally, when reacting Ca(NO3)2 with NaOH, a precipitate that is insoluble in excess of NaOH would be formed, while when reacting Zn(NO3)2 with NaOH, a precipitate would initially be formed, but it would dissolve in excess of NaOH.

Therefore, we could tell Ca(NO3)2 and Zn(NO3)2 apart from their reactions with KOH and Na2C2O4, based on the color of the precipitate formed and the speed which each precipitate is formed. Also, a reaction with NaOH could be used and the solubility of the formed precipitate in excess NaOH could also be used to differentiate Ca(NO3)2 and Zn(NO3)2.

Answer:

A) Using the ideal gas law, we can calculate the number of moles of gas present:

```

PV = nRT

```

where:

* P = pressure (atm) = 1 atm

* V = volume (L) = 5.6 L

* n = number of moles of gas

* R = ideal gas constant = 0.08206 L atm / mol K

* T = temperature (K) = 273.15 K

Solving for n, we get:

```

n = (P * V) / RT

```

```

n = (1 atm * 5.6 L) / (0.08206 L atm / mol K * 273.15 K)

```

```

n = 0.25 mol

```

Therefore, there are 0.25 moles of gas present.

B) The molar mass of the gas can be calculated by dividing the mass of the gas (8.0 g) by the number of moles of gas (0.25 mol):

```

Molar mass = Mass / n

```

```

Molar mass = 8.0 g / 0.25 mol

```

```

Molar mass = 32 g/mol

```

The molar mass of the gas is 32 g/mol.

C) The common atmospheric gas with a molar mass of 32 g/mol is oxygen (O2). Therefore, the gas that was collected is oxygen.

Explanation:

150 grams of NaOH is approximately equal to 2.256 x 10^24 particles of NaOH.

To convert grams of NaOH to particles of NaOH, we need to use the concept of molar mass and Avogadro's number. The molar mass of NaOH is calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H) together. It can be determined as follows:

Na: 22.99 g/mol

O: 16.00 g/mol

H: 1.01 g/mol

Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol

Now, we can use the molar mass to convert grams of NaOH to moles. Since 1 mole contains Avogadro's number (approximately 6.022 x 10^23) particles, we can determine the number of particles as follows:

150 g NaOH * (1 mol NaOH / 40.00 g NaOH) * (6.022 x 10^23 particles / 1 mol NaOH) ≈ 2.256 x 10^24 particles

It's important to note that this calculation assumes the substance is pure NaOH and that the molar mass and Avogadro's number are accurate.

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Answer:

All the above I believe

Explanation:

Answer:

all of above!

Explanation:

Im pretty sure, because all of them happen when forming a sedimentary rock.

Answer:

awdaef

Explanation:

Answer:

Answer is A

Areas at high elevations have cooler temperatures.

I presume you're doing earth science B?

Answer: It is A) Areas at high elevations have cooler temperatures.

Explanation: I got it correct.

The Friedel-Crafts acylation of bromobenzene using the acylium ion Formed in (a) proceeds via the following mechanism:

Formation of the acylium ion:

RCOCl + AlCl3 → RCO+ + AlCl4-

Coordination of the acylium ion with AlCl3:

RCO+ + AlCl3 → [RCO-AlCl3]+

Attack of the ortho carbon of bromobenzene on the electrophilic carbon of the acylium ion, forming a resonance-stabilized carbocation intermediate:

[RCO-AlCl3]+ + BrC6H4H → [RCO-C6H4-Br]+ + AlCl4-

Deprotonation of the carbocation intermediate by AlCl4-, forming the final product:

[RCO-C6H4-Br]+ + AlCl4- → RCO-C6H4-Br + AlCl3

Overall reaction:

RCOCl + BrC6H5 + AlCl3 → RCO-C6H4-Br + AlCl3

In this mechanism, the ortho carbon of bromobenzene acts as a nucleophile, attacking the electrophilic carbon of the acylium ion. The coordination of the acylium ion with AlCl3 helps to stabilize the positive charge on the carbocation intermediate. The final product is formed through deprotonation of the carbocation intermediate by AlCl4-

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Answer:

water

Explanation:

specific heat means the quantity of heat required to raise the temperature of one gram of a substance by one Celsius degree.

we would need 2625.13 grams (or 2.62513 kilograms) of NaCl.

To calculate the mass of NaCl required to produce a 26.4 mol/L solution with a 1.7 L volume, we need to use the formula that relates the mass of solute, moles of solute, and molarity:Molarity (M) = moles of solute / liters of solution Rearranging this formula, we get:moles of solute = Molarity (M) x liters of solutionWe can use this formula to find the moles of NaCl needed:moles of NaCl = 26.4 mol/L x 1.7 L = 44.88 molNow, we can use the molar mass of NaCl to convert from moles to grams. The molar mass of NaCl is 58.44 g/mol:mass of NaCl = moles of NaCl x molar mass of NaClmass of NaCl = 44.88 mol x 58.44 g/mol = 2625.13 gTo produce a 26.4 mol/L solution with a 1.7 L volume.

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Newton's first law states that unless acted on by an unbalanced force, an object will remain at rest or move at a constant speed in a straight line. To move the book on your desk, Newton's first law requires an unbalanced force.

The First Law of Motion of Newton (Inertia) Unless acted on by an unbalanced force, an object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line.

When an unbalanced force acts on an object, the object's motion changes. The object's speed or direction may change. Friction is a force that opposes motion or the tendency for motion between two objects in contact.

Thus, an object will remain at rest or move at a constant speed in a straight line.

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The general gas equation, commonly referred to as the ideal gas law, represents the state of a fictitious ideal gas through an equation. Here the mass of helium gas required to pressurize 86 L tank to 201 atm is 2561.8 g.

According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.

The ideal gas equation is given as:

PV = nRT

n = PV / RT

56°C = 329 K

n = 201 × 86 / 0.08206 × 329 = 640.45 mol

Molar mass of 'He' = 4.00 g / mol

Mass =  640.45 × 4.00 = 2561.8 g

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The percent yield of the reaction if 15.68 g of nitrogen and 10.74 g of hydrogen react to form 5.24 g of ammonia would be 82.35%.

From the balanced equation of the reaction below:

\(3H_2(g) + N_2 < - - > 2NH_3.\)

The mole ratio of hydrogen to nitrogen is 3:1.

The molar mass of hydrogen is 2 g/mol

The molar mass of nitrogen is 28 g/mol

Recall that: mole = mass/molar mass

Mole of 10.74 g hydrogen = 10.74/2

                                           = 5.37 mol

Mole of 15.68 nitrogen = 15.68/28

                                      = 0.56 mol

This means hydrogen is in excess while nitrogen is the limiting reactant. The mole ratio of nitrogen and ammonia is 1:2. Thus, the equivalent mole of ammonia produced would be:

0.56 x 2 = 1.12 mol

Molar mass of ammonia = 17 g/mol

Mass of 1.12 mol ammonia = 1.12 x 17

                                            = 19.04 grams

Percent yield of the reaction = 15.68/19.04 x 100

                                               = 82.35%

In other words, the percent yield of the reaction is 82.35%.

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The desired reaction is 2Al(s) + 3CO2 from Al2O3(s) + 3CO(g) (g) The reactions include 2 Al(s), 3/2 O2(g), and Al2O3(s), with H = 1675.7kJ. ————————— (1) CO(g) = CO2 + 1/2 O2(g) (g).

As a result, the enthalpies of a reactants and products are added together, and the result is used to compute the enthalpy of a reaction. This endothermic process generates and absorbs environmental heat if H is positive. This reaction is exothermic so emits heat into the environment if H is negative.

A negative H indicates that heat is transferred from the a system towards its surroundings, whereas a positive H indicates that heat is transferred from the surroundings into the system. An enthalpy of reaction (Hrxn) for a chemical reaction is the difference of enthalpy between the products and reactants; Hrxn is measured in kilojoules per mole.

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Two substituents are taken out of the reactant molecule in an elimination reaction to create the product.

In a one- or two-step method, two substituents are eliminated from a molecule in an organic reaction known as an elimination reaction. Atoms are eliminated as molecules and compounds during an elimination reaction. Two substituents are taken out of the reactant molecule in an elimination reaction to create the product. A metal, an acid, or a base typically catalyze elimination. E1 and E2 reactions are the two primary categories of elimination processes in organic chemistry. Both E1 and E2 processes are also known as alcohol elimination reactions and alkyl halide elimination reactions, respectively. Any organic chemical reaction known as an elimination reaction in which two atoms and groups of atoms are eliminated.

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Answer: The mass of \(CO_2\) and \(H_2O\) produced are 336.6 g and 183.6 g respectively.

Explanation:

The combustion reaction between propane and oxygen leads to formation of carbon dioxide and water.

Law of Conservation of mass states that the mass will remain constant for a balanced equation. This is carried out when the total number of atoms on reactant side is same as the total number of atoms on the product side. Thus the equation must be balanced.

\(C_3H_8+5O_2\rightarrow 3CO_2+4H_2O\)

a) 1 mol of propane produces = 3 moles of \(CO_2\)

Thus 2.55 mol of propane produces = \(\frac{3}{1}\times 2.55=7.65 moles of \(CO_2\)

mass of \(CO_2=moles\times {\text {molar mass}}=7.65mol\times 44g/mol=336.6g\)

b) 1 mol of propane produces = 4 moles of \(H_2O\)

Thus 2.55 mol of propane produces = \(\frac{4}{1}\times 2.55=10.2 moles of \(H_2O\)

mass of \(H_2O=moles\times {\text {molar mass}}=10.2mol\times 18g/mol=183.6g\)

The mass of \(CO_2\) and \(H_2O\) produced are 336.6 g and 183.6 g respectively.

Hydronium ion concentration, [H3O+], in a solution of pH = 1.76 is

[H3O+] = 0.01737800828M = 0.02M

pH is calculated using the expression :

\(Therefore : H_3O^+=10^(-pH)\)

\(Thus, H_3O^+=10^(-1.76) = 0.02M\)

Therefore, there are 0.02 moles of hydronium (H3O+ ) ions present in each liter of the solution.

Or, the concentration of the solution is about 0.02 M.

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Answer:

the answer to the answer above is motion

Answer: The balanced chemical equation for the reaction between potassium iodide (KI) and calcium chloride (CaCl₂) in aqueous solution can be written as:

2KI(aq) + CaCl₂(aq) → CaI₂(aq) + 2KCl(aq)

In this reaction, two molecules of potassium iodide react with one molecule of calcium chloride to produce one molecule of calcium iodide and two molecules of potassium chloride. The equation is balanced because the number of atoms of each element is equal on both sides of the equation.

Explanation: Please mark as brainliest for me love! <3

Answer:

A - Liquid molecules forming a gas and gas molecules forming a

liquid are equal in number.

Explanation:

A P E X

Answer:

There is 13.48 grams of P2O5 formed

Explanation:

Step 1: Data given

A decomposition of a sample of diphosphorus trioxide forms 1.29 g phosphorus to every 1.00 g oxygen.

Mass of P = 5.89 grams

Molar mass of O2 = 32.0 g/mol

atomic mass of P = 30.97 g/mol

molar mass of P2O5 = 141.94 g/mol

Step 2: The balanced equation

4P(s)+5O2(g)⇔ 2P2O5(s)

Step 3: Calculate moles of P

Moles P = Mass P / atomic mass P

Moles P = 5.89 grams / 30.97 g/mol

Moles P = 0.190 moles

Step 4: Calculate moles of P2O5

For 4 moles P we need 5 moles O2 to produce 2 moles P2O5

For 0.190 moles of P we'll have 0.190/2 = 0.095 moles P2O5

Step 5: Calculate mass of P2O5

Mass P2O5 = moles P2O5 * molar mass P2O5

Mass P2O5 = 0.095 moles * 141.94 g/mol

Mass P2O5 = 13.48 grams

There is 13.48 grams of P2O5 formed

Answer: 78.59 lb/ft

Explanation:

use stoichiometry to solve

1.26g/cm x ( 1pound/ 454 grams) X (28317 Cm/1 ft) =  78.59 pounds/ft

Answer:

Go to the pool , Bake, sleep, Watch ,Movies, go out with my freinds<3

Explanation:

 Mass of gas produced : 15 g

In general, the reaction takes place in the open air. If the reaction results in the form of gas as in combustion, the mass of the reaction results will be smaller than the original mass.  

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

So In a closed/isolated system, the total mass of the substance before the reaction will be equal to the total mass of the reaction product.

Mass of solution A  = 70 g

Mass of solution B = 35 g

Mass of mixture : 90 g

Mass of gas : X

Mass of gas X :

\(\tt mass~reactants(mass~A+mass~B)-mass~products(mass~mixture+mass~gas)\\\\mass~gas=(70+35)-90\\\\mass~gas=15~g\)

Answer:

D

Explanation:

Your supervisor is in charge of you so therefore you must always tell your supervisor first.

Answer:

C.

Explanation:

Because you might get hurt, could be dangerous.

Answer:

The correct answer is - 46.60 mL.

Explanation:

To find the volume of the gas at its new increased temperature we need to use Charl Law that shows the direct relationship between Volume and Temperature while Pressure remains constant.

V1 = 40 ml

T1  = 30 degree C + 273 = 303 K

V2  = ?

T2  = 80 degree C + 273 = 353 K

Charl Equation is:

V 1/T 1  =  V 2/ T 2

(V1) * (T2)/ T1= V2

placing value:

40*353/303 = V2

= 14120/303

Vf = 46.60 mL

The volume (in mL) of the 0.167 M sodium hydroxide, NaOH solution needed is 26.9 mL

The balanced equation for the reaction is given below:

HClO₄ + NaOH —> NaClO₄ + H₂O

The volume of NaOH can be obtained as follow:

MaVa / MbVb = nA / nB

(0.242 × 18.53) / (0.167 × Vb) = 1

4.48426 / (0.167 × Vb) = 1

Cross multiply

0.167 × Vb = 4.48426

Divide both side by 0.167

Vb = 4.48426 / 0.167

Vb = 26.9 mL

Thus, the volume of the NaOH solution needed is 26.9 mL

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Answer:The oxygen molecules are more loaded and heavier which weighs down and the molecules and don't allow the same amount of moving ability as the hydrogen moles have.

Explanation:

CuI2 is not a known compound. Copper compounds typically have different oxidation states for copper, resulting in various compound names.

Copper(II) oxide (CuO): It is a black solid compound where copper is in the +2 oxidation state. It is commonly used as a pigment and in catalytic reactions.

Copper(II) sulfate (CuSO4): It is a blue crystalline compound in which copper is in the +2 oxidation state. It is used in various applications such as agriculture, electroplating, and as a laboratory reagent.

Copper(I) oxide (Cu2O): It is a red crystalline compound in which copper is in the +1 oxidation state. It is used as a pigment, in solar cells, and as a catalyst.

Copper(II) chloride (CuCl2): It is a greenish-brown solid compound in which copper is in the +2 oxidation state. It is utilized in various chemical processes, including etching and catalyst synthesis.

Copper(II) nitrate (Cu(NO3)2): It is a blue crystalline compound where copper is in the +2 oxidation state. It is commonly used in the production of catalysts, as a coloring agent, and in electroplating.

These are just a few examples of copper compounds with different oxidation states and properties. It's important to note that the compound CuI2 mentioned in the question, if it exists, would be an exception to the typical nomenclature for copper compounds.

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