At a constant temperature of 30 °C, an ideal gas occupies 2.78 Liters at a pressure of 1.27 atm. What will be the volume (L) at a pressure of 3.95 atm?

Answer:

0.189 g

Explanation:

Step 1: Write the balanced equation

NCl₃ + 3 H₂O ⇒ NH₃ + 3 HCIO

Step 2: Calculate the moles corresponding to 1.33 g of NCl₃

The molar mass of NCl₃ is 120.36 g/mol.

1.33 g × 1 mol/120.36 g = 0.0111 mol

Step 3: Calculate the moles of NH₃ produced from 0.0111 moles of NCl₃

The molar ratio of NCl₃ to NH₃ is 1:1. The moles of NH₃ produced are 1/1 × 0.0111 mol = 0.0111 mol.

Step 4: Calculate the mass corresponding to 0.0111 moles of NH₃

The molar mass of NH₃ is 17.03 g/mol.

0.0111 mol × 17.03 g/mol = 0.189 g

Answer:

12

Explanation:

nNaCl= 4x3=12

Answer:

I think A should be the answer because oxygen is the chemical change of carbon.

Answer:

Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.

CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________

Explanation:

Given alcohol is propanol.

When it reacts with TsCl, the hydrogen in -OH group is replaced with tosyl group.

Pyridine is a weak base and it neutralizes the HCl (acid) formed during the reaction.

The reaction is shown below:

Answer:

1

Explanation:

1

Answer:

D. the minerals they contain

Hope this answer is right!!

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

[tex]rate = k[P]^{2} [Q][/tex]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

[tex]rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }[/tex]

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

[tex]k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4[/tex]

Second row:

2. Rate value:

[tex]rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1[/tex]

3.Third row:

[tex][Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}[/tex]

4. Fourth row:

[tex][P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}[/tex]

Answer: The mass of [tex]CO_2[/tex] produced is 12.32 g

Explanation:

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

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

Given mass of ethane = 4.21 g

Molar mass of ethane = 30 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol[/tex]

Given mass of oxygen gas = 31.9 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol[/tex]

The chemical equation for the combustion of ethane follows:

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

By stoichiometry of the reaction:

If 2 moles of ethane reacts with 7 moles of oxygen gas  

So, 0.140 moles of ethane will react with = [tex]\frac{7}{2}\times 0.140=0.49mol[/tex] of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, ethane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of ethane produces 4 moles of [tex]CO_2[/tex]

So, 0.140 moles of ethane will produce = [tex]\frac{4}{2}\times 0.140=0.28mol[/tex] of [tex]CO_2[/tex]

We know, molar mass of [tex]CO_2[/tex] = 44 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g[/tex]

Hence, the mass of [tex]CO_2[/tex] produced is 12.32 g

Answer:

For gases such as hydrogen, oxygen, nitrogen, helium, or neon, deviations from the ideal gas law are less than 0.1 percent at room temperature and atmospheric pressure. Other gases, such as carbon dioxide or ammonia, have stronger intermolecular forces and consequently greater deviation from ideality.

Explanation:

Answer:

Option 1. NO

Explanation:

The balanced equation for the reaction is given below below:

4NH₃ + 6NO —> 5N₂ + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Therefore, 1.24 moles of NH₃ will react with = (1.24 × 6)/4 = 1.86 moles of NO

From the calculation made above, we can see that a higher amount of NO (i.e 1.86 moles) than what was given (i.e 1.79 moles) is needed to react completely with 1.24 moles of NH₃.

Therefore, NO is the limiting reactant and NH₃ is the excess reactant.

Thus, the 1st option gives the correct answer to the question

Answer:

1. NO .

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the limiting reactant by simply calculating the moles of any product, say N2, via the moles of each reactant and including the corresponding mole ratio (4:5 and 6:5):

[tex]1.24molNH_3*\frac{5molN_2}{4molNH_3}=1.55molN_2 \\\\1.79molNO*\frac{5molN_2}{6molNO}=1.50molN_2[/tex]

Thus, since NO yields the fewest moles of N2 product, we infer it is the limiting reactant.

Regards!

Answer:

95.9 kg

Explanation:

First we convert 15.0 mi² to m²:

Then we convert 27.0 ft to m:

Now we calculate the total volume of the lake:

Converting 3.20x10⁸ m³ to L:

Now we calculate the total mass of mercury in the lake, using the given concentration:

Finally we convert μg to kg:

Answer:

Group 1A: alkali metals, or lithium family.

Group 2A: alkaline earth metals, or beryllium family.

Group 7A: the manganese family.

Group 8A: the iron family.

Explanation:

Answer:

1A: Alkali Metals

2A: Alkaline Earth Metals

7A: Halogens

8A: Noble Gases

Answer:

Easy my dude let me help you out

A.In

B.27

C.73

D.49

E.56

F.56

G.114

H.180

Also with protons and electrons they equal the same atomic number

Answer:

1. Both

2. Phosphorylation

3. Both

4. Phosphorylation

5. Oxidative.

6. Both

Explanation:

Phosphorylation only occurs in chloroplast and it involves larger electrical component. Both Phosphorylation and oxidative occurs in mitochondria and it involves proton gradient. They occur in plants to produce ATP. Oxidative involves in smaller electrical component.

Photophosphorylation is a process that captures the solar energy from the sun to transform it into chemical energy. It occurs in the chloroplast of a plant cell.

Photophosphorylation is a process of converting solar energy from the sun to ATP needed by plants and other organisms for cellular function and activity. This process takes place in the chloroplast of the plant cell and requires electrical components.

Oxidative Phosphorylation is the process of producing ATP with the help of oxygen and enzymes hence, occurs in aerobic cells. It does not need a larger electrical component.

Both phosphorylation and oxidative phosphorylation occurs in the mitochondria of plants cells and involves a proton gradient for the formation of ATP.

Therefore, oxidative phosphorylation option 5. involves a smaller electrical component, phosphorylation option 2. occurs in the chloroplast, and option 4. needs a larger electrical component.

Learn more about phosphorylation here:

https://brainly.com/question/1870229

Answer:

36g

Explanation:

you need to know the equation mass=moles*mr (in this case mr of carbon which is 12)

so 3*12=36g

hope this helps :)

Answer:

NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)

Explanation:

Sodium hydrogen sulfate is a strong electrolyte, that is, when dissolved in water it completely dissociates into the cation sodium and the anion hydrogen sulfate. The corresponding chemical equation is:

NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)

The question is incomplete, the complete question is:

When aqueous solutions of NaCl and [tex]Pb(NO_3)_2[/tex] are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of

Answer: The mass of lead chloride produced is 1.96 g

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}[/tex] .....(1)

Given values:

Molarity of NaCl = 0.1000 M

Volume of the solution = 140.7 mL

Putting values in equation 1, we get:

[tex]0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol[/tex]

The chemical equation for the reaction of NaCl and lead nitrate follows:

[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]

By the stoichiometry of the reaction:

If 2 moles of NaCl produces 1 mole of lead chloride

So, 0.01407 moles of NaCl will produce = [tex]\frac{1}{2}\times 0.01407=0.007035mol[/tex] of lead chloride

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

The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)

Molar mass of lead chloride = 278.1 g/mol

Plugging values in equation 2:

[tex]\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g[/tex]

Hence, the mass of lead chloride produced is 1.96 g

Answer:

3.59x10⁻⁴ mol

Explanation:

Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:

We input the data given by the problem:

And solve for n:

Answer:

the answer is b.CH3NO2 I guess I'm correct

Answer: 26 moles of [tex]PH_3[/tex] are required for the reaction.

Explanation:

We are given:

Moles of [tex]P_4O_{10}[/tex] = 6.5 moles

The given chemical reaction follows:

[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]

By the stoichiometry of the reaction:

If 1 mole of [tex]P_4O_{10}[/tex] is produced by 4 moles of [tex]PH_3[/tex]

So, 6.5 moles of [tex]P_4O_{10}[/tex] will be produced by = [tex]\frac{4}{1}\times 6.5=26mol[/tex] of [tex]PH_3[/tex]

Hence, 26 moles of [tex]PH_3[/tex] are required for the reaction.

Answer:

B

Explanation:

So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.

Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

Solution :

The equation is :

[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15  mL = 40 mL

The pH of the solution is 6.50

Calculating the [tex]K_a[/tex] of HA

[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

Let s calculate the concentration of HA and NaOH

[tex]$[HA] = \frac{^nH_A}{V}$[/tex]

        [tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]

       = 0.07125 M

[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]

            [tex]$=\frac{0.001425 mol}{0.04L}$[/tex]

           = 0.0356 M

                                      [tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]

Initial conc. (M)            0.07125 M       0.0356 M            0 M

Change in conc. (M)   -0.0356 M       -0.0356 M        + 0.0356 M

Equilibrium conc. (M)   0.03565 M        0 M                0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]

Now for [tex][H^+][/tex]

[tex]$[H^+] = 10^{-pH}$[/tex]

       [tex]$=10^{-6.5}$[/tex]

       [tex]$=3.16 \times 10^{-7}$[/tex]

Calculating the value of [tex]K_a[/tex],

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

     [tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]

     [tex]$=3.16\times 10^{-7}$[/tex]

Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].

Answer:

The maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%

Note: The question is incomplete. A similar but complete question is given below:

The solubility of benzoic acid in water is 6.80g per 100mL at 100 degrees C and 0.34 g per 100mL at 25 degrees C.

Calculate the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water, assuming the solution is filtered at 25 degrees C.

Explanation:

Solubility of benzoic acid in water at 100 degrees C = 6.80g per 100mL

Solubility of benzoic acid in water  at 25 degrees C =  0.34 g per 100mL

Mass of benzoic acid to be theoretically recovered from 100 mL of water = 6.80 g - 0.34 g = 6.46 g

At 25 degrees;

0.34 g of benzoic acid is present in 100 mL of water

x g of  benzoic acid will be present in 15 mL of water

x = 0.34 × 15 / 100 = 0.051 g

Mass of benzoic acid to be theoretically recovered from 25 mL of water = 1.00 g - 0.051 g = 0.949 g

Maximum theoretical percent recovery = (mass recovered / original mass dissolved) x 100%

Maximum theoretical percent recovery =  (0.949 / 1.00) × 100% = 94.9 %

Therefore, the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%

Answer:

[tex]\%m=66.7\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.

Next, we apply the following equation to obtain the required concentration:

[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]

Regards!

Answer:

Explanation:

578kj/mol

Answer:

The pressure will be 933.33 Kpa

Explanation:

Given that:

Volume V₁ = 200 cm³  (note, there is a mistake in the volume. It is supposed to be 200 cm³)

Pressure P₁ = 700 Kpa

Pressure P₂ = ??? (unknown)

Volume V₂ = 150 cm³

Temperature = constant

Using Boyle's law:

PV = constant

i.e.

P₁V₁ = P₂V₂

700 Kpa × 200 cm³ = P₂ × 150 cm³

P₂ = (700 Kpa × 200 cm³)/150 cm³

P₂ = 933.33 Kpa