Many ski resorts and mountain cities can be reached directly by planes which deposit travelers a mile or more above sea level. This can result in altitude sickness due to hypoxemia, or reduced oxygen in the blood, caused by the unaccustomed exposure to the lower atmospheric pressure at high elevations. On a given day, the prevailing atmospheric pressure in Yellowstone, MT (elevation 2031 meters) might be 0.775 atm. Calculate this pressure expressed in mmHg and in torr.

Answer:

(a): The expression of equilibrium constant is [tex]K_{eq}=\frac{[NO]^2}{[N_2][O_2]}[/tex]

(b): The equation to solve the concentration of NO is [tex][NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}[/tex]

(c): The concentration of NO is 0.0017 M.

Explanation:

The equilibrium constant is defined as the ratio of the concentration of products to the concentration of reactants raised to the power of the stoichiometric coefficient of each. It is represented by the term [tex]K_{eq}[/tex]

(a):

The given chemical equation follows:

[tex]N_2+O_2\rightarrow 2NO[/tex]

The expression for equilbrium constant will be:

[tex]K_{eq}=\frac{[NO]^2}{[N_2][O_2]}[/tex]

(b):

The equation to solve the concentration of NO follows:

[tex][NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}[/tex]            ......(1)

(c):

Given values:

[tex]K_{eq}=1.2\times 10^{-4}[/tex]

[tex][N_2]_{eq}=0.166M[/tex]

[tex][O_2]_{eq}=0.145M[/tex]

Plugging values in equation 1, we get:

[tex][NO]=\sqrt{(1.2\times 10^{-4})\times 0.166\times 0.145}[/tex]

[tex][NO]=\sqrt{2.88\times 10^{-6}}[/tex]

[tex][NO]=0.0017 M[/tex]

Hence, the concentration of NO is 0.0017 M.

Answer:

93.4 mL

Explanation:

Let's state the reaction:

2AgNO₃ +  Na₂CO₃  →   Ag₂CO₃  +  2NaNO₃

We determine the moles of sodium carbonate:

0.3572 g . 1mol / 105.98g = 3.37×10⁻³ moles

Ratio is 1:2. We say:

1 mol of sodium carbonate react to 2 moles of silver nitrate

Then, our 3.37×10⁻³ moles of carbonate may react to: 3.37×10⁻³ . 2

= 6.74×10⁻³ moles

If we convert to mmoles → 6.74×10⁻³ mol . 1000 mmol / mol = 6.74 mmol

Molarity is mol/L but we can use mmol /mL

6.74 mol / volume in mL = 0.07216 M

6.74 mol / 0.07216 M = volume in mL → 93.4 mL

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

[tex]N_2(g)+3F_2(g)\rightarrow 2NF_3(g)[/tex]

A sample contains 19.3 g of [tex]N_2[/tex] is reacted with 19.3 g of [tex]F_2[/tex]. Now we need to find the amount of [tex]NF_3[/tex] that can be formed by the complete reactions of each of the reactants.

If all of the [tex]N_2[/tex] was used up in the reaction, how many moles of [tex]NF_3[/tex] would be produced?

Answer: 1.378 moles of [tex]NF_3[/tex] are produced in the reaction.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]       ......(1)

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, [tex]N_2[/tex] is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of [tex]N_2[/tex] = 19.3 g

Molar mass of [tex]N_2[/tex] = 28.02 g/mol

Putting values in equation 1:

[tex]\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol[/tex]

For the given chemical reaction:

[tex]N_2(g)+3F_2(g)\rightarrow 2NF_3(g)[/tex]

By the stoichiometry of the reaction:

1 mole of [tex]N_2[/tex] produces 2 moles of [tex]NF_3[/tex]

So, 0.689 moles of [tex]N_2[/tex] will produce = [tex]\frac{2}{1}\times 0.689=1.378mol[/tex] of [tex]NF_3[/tex]

Hence, 1.378 moles of [tex]NF_3[/tex] are produced in the reaction.

[tex]\huge\mathsf{\red{\underline{\underline{Answer}}}}[/tex]

[tex]{\green{\dashrightarrow}}[/tex]When you mix the rubbing alcohol with water, the latter's molecules make hydrogen bonds with the water molecules. The alcohol dissolves in the water to form a homogenous solution, so you cannot distinguish the alcohol and the water anymore.

Answer:

1.20 × 10³ torr

Explanation:

Step 1: Given data

Step 2: Calculate the final pressure of the gas

Considering the constant volume, if we assume the gas behaves ideally, we can calculate its final pressure using Gay-Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = 822 torr × 475 K/325 K = 1.20 × 10³ torr

Answer:

The elements present in calcium phosphate-based on molecular formula are calcium, phosphorus, and oxygen. The molecular mass of Ca or calcium is 40.08 grams per mole, the molecular mass of P or phosphorus is 30.97 grams per mole, and the molecular mass of oxygen is 16.00 grams per mole. So, in combination, the molecular mass of Ca₃(PO₄)₂ is, = (3 × 40.08 + 2 ×3 0.97 + 8 × 16) = 310.18 grams per mole

Now, the mass percent of calcium is, = n × molecular mass of calcium/ molecular mass of Ca₃(PO₄)₂ × 100% = 3 × 40.08 / 310.18 × 100 = 38.76 % Thus, the mass percent of the calcium in calcium phosphate is 38.76%.  

The mass percent of phosphorus (P), = n × molecular mass of phosphorus/molar mass of Ca₃(PO₄)₂ × 100% = 2 × 30.97/310.18 × 100 = 19.97% Thus, the mass percent of phosphorus in calcium phosphate is 19.97%.  

The mass percent of oxygen (O), = n × molecular mass of oxygen/molar mass of Ca₃(PO₄)₂ × 100% = 8 × 16/310.18 × 100% = 41.27% Thus, the mass percent of oxygen in calcium phosphate is 41.27%.

Answer:

[tex]\mathbf{FeCl_3 + AgNO_3 \to 3 AgCl+Fe(NO_3)_3}[/tex]

Explanation:

From the given question,

The chemical equation is:

[tex]\mathbf{FeCl_3 + AgNO_3 \to 3 AgCl+Fe(NO_3)_3}[/tex]

We are given that:

mass of [tex]FeCl_3[/tex] = 45 g

number of moles of [tex]FeCl_3[/tex]  = 45 g/162.2 g/mol

= 0.28 mol

mass of [tex]AgNO_3[/tex] = 25g

number of moles of [tex]AgNO_3[/tex]  = 25/169.87

= 0.147 mol

From the given equation 1 mole of [tex]FeCl_3[/tex] is required to make 3 mole of [tex]AgNO_3[/tex]  

0.28 mole of [tex]FeCl_3[/tex] = 0.28 × 3 = 0.84 mol of [tex]AgNO_3[/tex]  

Here [tex]AgNO_3[/tex]  is the limiting reagent.

Thus,

3 mole of [tex]AgNO_3[/tex]  = 3 moles of AgCl

0.147 mole of [tex]AgNO_3[/tex] = 0.147 mole of AgCl produced.

Answer:

the correct answer is 12 sides (:

Explanation:

Answer:

~4,317.5 moles of Silicon

Explanation:

Using Avogadro's contant we know that:

1mole = 6.022 x 10^23 atoms

So,to calculate the number of moles in 2.60 x 10^27 atoms of Silicon:

=(2.60 x 10^27 x 1)/(6.022 x 10^23)

~4,317.5 moles of Silicon

Hope it helps:)

Answer:

Q = 669.44 J

Explanation:

Given that,

Mass of water, m = 32 g

The temperature change from 25.0°C to 20.0°C.

We need to find the amount of heat energy transferred. Let it is Q. We know that,

[tex]Q=mc\Delta T[/tex]

Where

c is the specific heat of water

Put all the values,

[tex]Q=32\times 4.184 \times (20-25)\\Q=669.44\ J[/tex]

So, 669.44 J of heat energy is transferred from the water.

Answer:

0.871 mL

Explanation:

Step 1: Write the balanced neutralization reaction

KOH + HF ⇒ KF + H₂O

Step 2: Calculate the reacting moles of HF

9.55 mL of 0.114 M HF react.

9.55 × 10⁻³ L × 0.114 mol/L = 1.09 × 10⁻³ mol

Step 3: Calculate the moles of KOH needed to react with 1.09 × 10⁻³ moles of HF

The molar ratio of KOH to HF is 1:1. The moles of KOH needed are 1/1 × 1.09 × 10⁻³ mol = 1.09 × 10⁻³ mol.

Step 4: Calculate the volume of KOH solution that contains 1.09 × 10⁻³ moles of KOH

An unknown volume of 1.252 M KOH solution contains 1.09 × 10⁻³ moles of KOH.

1.09 × 10⁻³ mol × 1 L/1.252 mol = 8.71 × 10⁻⁴ L = 0.871 mL

Answer:

me too

Explanation:

Given :

Molarity of HCl solution, M = 0.686 M.

Number of moles, n = 0.037 moles.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

[tex]M = \dfrac{Number \ of \ moles \ of \ solute}{Volume \ in \ liter}\\\\0.686 = \dfrac{0.037}{V}\\\\V = \dfrac{0.037}{0.686}\ L\\\\V = 0.053935 \ L \\\\V = 53.94 \ ml[/tex]

Hence, this is the required solution.

Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...

Using an ionic chloride and Phosphoric acid

H3PO4 + NaCl ==> HCl + NaH2PO4

H3PO4 + NaI ==> HI + NaH2PO4

H2SO4 + NaCl ==> HCl + NaHSO4

This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.

The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.

2I-² === I2 + 2e-

Answer:

option number 4. 'coral'

Explanation:

-

Answer:

See explanation

Explanation:

Given the reaction;

3N2(g) + H2(g) = 2N3H(g).

We know that when a constraint is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.

In this case, when the pressure is increased eightfold, the equilibrium position will shift towards the direction where there is the least total volumes.

In this particular reaction, the forward reaction has the least total volumes and is favoured by an eightfold increase in pressure.

Hence, when the pressure is increased eightfold, more N3H(g) is obtained.

Answer:

The original concentration is "35 mg/ml".

Explanation:

According to the question,

The solution is diluted,

= 1:50

The initial volume,

V1 = 1 ml

Final concentration,

= 0.07 mg

then,

The final volume,

V2 = 500 ml

As we know,

⇒ [tex]V_1N_1=V_2N_2[/tex]

or,

⇒ [tex]N_1=\frac{V_2N_2}{V_1}[/tex]

On substituting the values, we get

⇒       [tex]=\frac{500\times 0.07}{1}[/tex]

⇒       [tex]=\frac{35}{1}[/tex]

⇒       [tex]=35 \ mg/ml[/tex]

A 0.07 mg/mL BHL solution was prepared by performing a 1:500 dilution on a 35 mg/mL solution.

A 1:500 dilution was carried out to prepare a 0.07 mg/mL BHL solution. We can find the concentration of the initial solution using the dilution rule.

It is an equation that relates the concentration and volume of a concentrated and a dilute solution. The mathematical expression is:

C₁ × V₁ = C₂ × V₂

where,

C₁ = C₂ × (V₂/V₁)

C₁ = 0.07 mg/mL × (500/1) = 35 mg/mL

A 0.07 mg/mL BHL solution was prepared by performing a 1:500 dilution on a 35 mg/mL solution.

Learn more about dilution here: https://brainly.com/question/1615979

Answer: The [tex]K_{sp}[/tex] value for calcium hydroxide at this temperature is [tex]1.08 \times 10^{-4}[/tex].

Explanation:

Given: Mass of [tex]Ca(OH)_{2}[/tex] = 0.225 g

Volume = 0.100 L

As moles is the mass of substance divided by its molar mass.

So, moles of [tex]Ca(OH)_{2}[/tex] (molar mass = 74 g/mol) is calculated as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{0.225 g}{74 g/mol}\\= 0.003 mol[/tex]

Molarity is the number of moles of substance present in a liter of solution.

Hence, molarity of given solution will be as follows.

[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.003 mol}{0.1 L}\\= 0.03 M[/tex]

The equation for dissociation of  [tex]Ca(OH)_{2}[/tex] is as follows.

[tex]Ca(OH)_{2} \rightarrow Ca^{2+} + 2OH^{-}[/tex]

This means that [tex][Ca^{2+}] = 0.03[/tex] and [tex][OH^{-}] = 2 \times 0.03 = 0.06[/tex]. Hence, [tex]K_{sp}[/tex] value for this reaction is calculated as follows.

[tex]K_{sp} = [Ca^{2+}][OH^{-}]^{2}\\= (0.03) \times (0.06)^{2}\\= 1.08 \times 10^{-4}[/tex]

Thus, we can conclude that the [tex]K_{sp}[/tex] value for calcium hydroxide at this temperature is [tex]1.08 \times 10^{-4}[/tex].

Answer:

John Dalton

Explanation:

Answer:

If you want to separate black grapes from the mixture of black and green grapes, then you will simply pick black grapes using your hands from the mixture. In this way you are actually using handpicking separation method.

Explanation:

Answer:

1.The substance(s) to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products

2.The substances that go into a chemical reaction are called the reactants, and the substances produced at the end of the reaction are known as the products

3.The law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. For example, the mass of sodium plus the mass of chlorine that reacts with the sodium equals the mass of the product sodium chloride.

Answer:

The difference between the three classes depends on where the force is, where the fulcrum is and where the load is. In a first class lever, the fulcrum is located between the input force and output force. In a second class lever, the output force is between the fulcrum and the input force. write the class of lever.

Explanation:

Answer:

5.56g of FeCl2 can be produced

Explanation:

To solve this question we must find the moles of each reactant. With the moles and the chemical equation we can find limiting reactant. With limiting reactant we can find the moles of FeCl2 and its mass as follows:

Moles HCl:

1.5L * (0.25mol / L) = 0.375 moles HCl

Moles Fe -Molar mass: 55.845g/mol-

2.45g * (1mol / 55.845g) = 0.0439 moles Fe

For a complete reaction of 0.375 moles HCl are needed:

0.375 moles HCl * (1mol Fe / 2mol HCl) = 0.1875 moles Fe

As there are just 0.0439 moles Fe, Fe is limiting reactant

1mol of Fe produce 1 mole of FeCl2, 0.0439 moles Fe produce 0.0439 moles of FeCl2. The mass is:

Mass FeCl2 -Molar mass: 126.751g/mol:

0.0439 moles Fe * (126.751g / mol) =

Answer:

Answer C

Explanation:

Got it off of q**zlet ;)

Answer

Its C confirmed

Explanation:

Just did the test

Answer:

Hi, there your answer is A. As the frequency of a wave increases, the shorter its wavelength is.

Explanation:

When frequency increases, wavelength decreases.

Hope this Helps :)