Is HNO3 an acid or a base



I NEED HELP ASAP​

Answer:

10573.9K

Explanation:

Using pressure law equation;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 46mmHg

P2 = 760mmHg

T1 = 640K

T2 = ?

Using P1/T1 = P2/T2

46/640 = 760/T2

Cross multiply

640 × 760 = 46 × T2

486400 = 46T2

T2 = 486400 ÷ 46

T2 = 10573.9K

We are given:

Initial concentration of HA: 0.200 Molar

The acid is 9.4% ionized

Dissociation constant (α) = (Percent Ionized) / 100 = 0.094

Molar concentration of H+ = 0.0188

Let's Chill! (making the ICE box):

Reaction:       HA         ⇄       H⁺      +      A⁻

Initial:           0.200M               -                -

Equilibrium: 0.200(1-α)      0.200α       0.200α

while we're here, let's confirm the given equilibrium concentration of H⁺ ions

from the table here, we can see that the equilibrium concentration of H⁺ ions is 0.200α, we know that α = 0.094

[H⁺] = 0.200α = 0.200 * 0.094 = 0.0188 M

which means that we're on the right track

We're basically scientists at this point (finding the dissociation constant):

Acid dissociation is nothing but the equilibrium constant, but for the dissociation of Acids

From the reaction above, we can write the equation of the acid dissociation constant:

Ka = [H⁺][A⁻] / [HA]

now, let's take the values from the 'equilibrium' row of the ice box the plug those in this equation

Ka = (0.200α)(0.200α) / [0.200(1-α)]

Ka = (0.200α)²/[0.200(1-α)]

plugging the value of α

Ka = (0.200*0.094)² / [0.200(0.906)]

Ka = (0.0188)² / 0.1812

Ka = 1.95 * 10⁻³

Light energy travels in the form of waves.

Answer:

How to Formulate an Effective Research Hypothesis

State the problem that you are trying to solve. Make sure that the hypothesis clearly defines the topic and the focus of the experiment.

Try to write the hypothesis as an if-then statement. ...

Define the variables.

Explanation:

Answer:

Number of moles  of sodium dissolved =  6.0 *10^23

Explanation:

The image for the question is attached

Solution

a) Total 181 ions of Na are dissolved

b)

The number of moles of sodium dissolved = 181/6.023 *10^23

Number of moles  of sodium dissolved = 5.987 * 10^23

Number of moles  of sodium dissolved =  6.0 *10^23

Answer:

In order from left to right, 7 (gamma), 5 (ultraviolet, now continue pattern), 4, 6, 2, 3, 1.

Answer:

Explanation:

Short answer: Yes.

The coefficients may not be conserved, but mass always has to be. Take this equation as an example

2 Mg3P2 ===>   6Mg + P4

There is a 2 on the left side and 6 and 1 on the right. I hope you mean that the coefficient 2 is not equal to 7.

But let's look a little closer. You have to look at the molecular structure of the left and right side.

2Mg3P2 has 6 Mgs and 4 Ps on the left side.

6Mg is on the right. They are free standing.

P4    has 1 molecule consisting of 4 Ps.

Everything balances.

This is a terrific question to be asking. You need to understand the internal balance numbers vs the molecular ones on the out side.

That sounds like a bit of gobbledygook and it takes a bit of study.

2 Mg3P2 means that Mg3P2 is written twice.

Mg3P2 ==> "3 Mg2+   +  2P3+   and there is another one written the same way.

Mg3P2 ==> "3 Mg2+   +  2P3+  

Answer:

I think you forgot to post the question or picture

Answer:

The correct answer is option 4, that is, there is exactly enough sodium carbonate.

Explanation:

Based on the given question, the reaction will be,

2 HCl (aq) + Na2CO3 (s) ⇒ 2 NaCl (aq) + CO2 (g) + H2O (l)

Therefore, for neutralizing 2 moles of HCl, one mole of Na2CO3 is required.

No of moles present in 1 Kg or 1000 grams of Na2CO3 will be,

Moles = Weight/Molecular mass of Na2CO3

Moles = 1000 / 106 = 9.43

Thus, 9.43 moles of Na2CO3 is present.

No of moles present in 1 liter of 9.75 M HCl is 9.75.

No. of moles present in 5 Liters of HCl (9.75 M),

= 5 × 9.75 = 48.75

Thus, for 2 moles of HCl 1 mole of Na2CO3 is required. Now for 48.75 moles of HCl, the moles required of Na2CO3 is 9.75. Therefore, for complete neutralization, the moles of Na2CO3 required is 9.75, and the present moles is 9.43.

Hence, there is exactly enough sodium carbonate.

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the chemical reaction between NaOH and HCl:

[tex]NaOH+HCl\rightarrow NaCl+H_2O[/tex]

Thus, since they react in a 1:1 mole ratio; we can now calculate the moles of each substance by using their volumes and molarities:

[tex]n_{NaOH}=0.0250L*0.81mol/L=0.02025molNaOH\\\\n_{HCl}=0.0650L*0.33mol/L=0.02145molHCl[/tex]

Now, since NaOH is in a fewer proportion, we infer just 0.02025 moles of HCl are consumed so that 0.0012 moles of this acid remain unreacted; in such a way, we infer that the NaOH is the limiting reactant for this reaction.

Regards!

Answer:

[tex](NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)[/tex]

Explanation:

Hello there!

In this case, according to the given information, we can set up the appropriate chemical equation when ammonium phosphate reacts with sodium chloride in aqueous solution:

[tex](NH_4)_3PO_4(aq)+NaCl(aq)\rightarrow NH_4Cl(aq)+Na_3PO_4(aq)[/tex]

Which stands for a double replacement reaction, whereby ammonium changes phosphate to chloride and sodium changes chloride to phosphate on the products side. In addition, we can balance the aforementioned equation as shown below:

[tex](NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)[/tex]

Regards!

Answer:

Explanation:

mass = 400 grams * [1 kg/1000 grams] = 0.400 kg

c = 387 Joules / (oC * kg)

Δt = 55 - 30 = 25 oC

E = m*c * Δt

E = 0.4 * 387 * 25

E = 3870 Joules

Answer:

Explanation:

photosynthesis

the given chemical reaction is photosynthesis.

During photosynthesis carbon dioxide absorbed by plants reacts with water in presence of sunlight to give glucose and oxygen.

Answer:

lololol

Explanation:

Answer:

Molarity = 0.809 M

mole fraction = 0.047

Explanation:

The complete question is

Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution -

Solution for molarity:

1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

1)  

Mass of 1.09 mole of acetone

= 1.09  mol x 58.0794 g/mol = 63.306 g

Density of acetone = 0.788 g/cm3  

Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

For ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

a) Molarity:

1.09 mol / 1.347 L = 0.809 M

Mole Fraction  

a) moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

b) moles of acetone:

1.09 / (1.09 + 21.71) = 0.047

Answer:

1. Empirical formula => C₂H₃O

2. Molecular formula => C₆H₉O₃

Explanation:

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

Mass of compound = 50 g

Mass of Carbon = 24.66 g

Mass of Hydrogen = 3.43 g

Molecular weight of compound = 146.0 amu

Empirical formula =?

Molecular formula =?

Next, we shall determine the mass of oxygen in the compound. This can be obtained as follow:

Mass of compound = 50 g

Mass of C = 24.66 g

Mass of H = 3.43 g

Mass of O =?

Mass of O = mass of compound – ( mass of C + mass of H)

= 50 – (24.66 + 3.43)

= 50 – 28.09

= 21.91 g

1. Determination of the empirical formula.

Mass of C = 24.66 g

Mass of H = 3.43 g

Mass of O = 21.91 g

Divide by their molar mass

C = 24.66 / 12 = 2.055

H = 3.43 / 1 = 3.43

O = 21.91 / 16 = 1.369

Divide by the smallest

C = 2.055 / 1.369 = 2

H = 3.43 / 1.369 = 3

O = 1.369 / 1.369 = 1

Therefore, the empirical formula of the compound is C₂H₃O

2. Determination of the molecular formula.

Molecular weight of compound = 146.0 amu

Empirical formula => C₂H₃O

Molecular formula =?

Molecular formula = [C₂H₃O]ₙ = molecular weight

Thus,

[C₂H₃O]ₙ = 146

[(12×2) + (3×1) + 16]n = 146

[24 + 3 + 16]n = 146

43n = 146

Divide both side by 43

n = 146 / 43

n = 3

Molecular formula = [C₂H₃O]ₙ

Molecular formula = [C₂H₃O]₃

Molecular formula = C₆H₉O₃

Answer:

That should be Hydrogen.

So The second option Is legit!

above because its above

Answer:

The explanation according to the given question is summarized below.

Explanation:

Given:

Tris,

= 250 mM

Glycine,

= 1.92 M

According to the solution,

For the dilution pf 10X to 1X buffer, we get

= [tex]1 \ ml \ of \ 10X \ buffer +9 \ ml \ of \ distilled \ water[/tex]

= [tex]10[/tex]

i.e.,

⇒ [tex]10X \ to \ 1X=1:10 \ dilution[/tex]

Now,

⇒ [tex]10X (250 \ mM\ Tris \ HCl, 1.92M\ Glycine, and\ 1 \ percent (\frac{w}{v} ) SDS) ---->1X(25 \ mM \ Tris \ HCl,0.193 M\ Glycine, and \ 0.1 \ percent(\frac{w}{v} )SDS)[/tex]

Answer:

non polar. polar ionic substance

Answer:

1.932 kg

Explanation:

First we convert 100.0 mL to L:

Then we calculate the mass of the substance, using the definition of density:

As the multiplication involves two numbers of 4 significant figures each, the answer needs to have 4 significants figures as well.

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

[tex]{\green{\dashrightarrow}}[/tex]A chemical compound is a chemical substance that is made of two or more atoms of different elements that share a chemical bond.

[tex]{\green{\dashrightarrow}}[/tex]A chemical formula represents the ratio of atoms per element that make up the chemical compound.

[tex]\large{\pink{\sf{5~ Examples~ of~ Compound~ are:-}}}[/tex]

Answer:

The amount after three year is 617934.1302

Explanation:

Complete question

A person places $530,000 in an investment account earning an annual rate of 5.25%, compounded continuously. Using the formula V = Pe^{rt}V=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 3 years

Solution

The formula for calculating compound interest is

[tex]A = p (1 + \frac{r}{n})^{nt}[/tex]

Substituting the given values we get -

[tex]A = 530,000 (1 + \frac{5.25}{100})^3\\A = 530,000 * ( 1+ 0.0525)^3\\A = 530,000 * ( 1.0525)^3\\A = 617934.1302[/tex]

The amount after three year is 617934.1302

Answer:

[tex]\boxed {\boxed {\sf 0.255 \ mol \ C }}[/tex]

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

Set up a ratio using the molar mass.

[tex]\frac {12.011 \ g \ C}{ 1 \ mol \ C}[/tex]

Since we are converting 3.06 grams to moles, we multiply by that value.

[tex]3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}[/tex]

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

[tex]3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}[/tex]                      

[tex]3.06 * \frac{1 \ mol \ C}{12.011 }[/tex]    

[tex]\frac {3.06}{12.011 } \ mol \ C[/tex]                                

[tex]0.25476646 \ mol \ C[/tex]

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

[tex]0.255 \ mol \ C[/tex]

3.06 grams of carbon is approximately 0.255 moles of carbon.

The question is incomplete. The complete question is :

C. Balance these fossil-fuel combustion reactions. (1 point)

C8H18(g) + 12.5O2(g) →    ____CO2(g) + 9H2O(g) + heat

CH4(g) + ____O2(g) →     ____CO2(g) + ____H2O(g) + heat

C3H8(g) + ____O2(g) →    ____CO2(g) + ____H2O(g) + heat

C6H6(g) + ____O2(g)  →   ____CO2(g) + ____H2O(g) + heat

Solution :

C8H18(g) + 12.5O2(g) →    __8__CO2(g) + 9H2O(g) + heat

When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.

CH4(g) + __2__O2(g) →     __1__CO2(g) + __2__H2O(g) + heat

When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.

C3H8(g) + __5__O2(g) →    __3__CO2(g) + __4__H2O(g) + heat

When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.

C6H6(g) + __1/2__O2(g)  →   __6__CO2(g) + __3__H2O(g) + heat

When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.