The total pressure in a container is 37.9 psi. There are two gases in the container. Gas A exerts a pressure of 18.4 psi. What is the pressure of gas B?

Answer:

we cant draw jn this

Explanation:

because it contin citric acid

Answer:

49.1 g

Explanation:

First we convert 9.67x10²³ atoms of oxygen into moles, using Avogadro's number:

Then we calculate how many P₃O₉ moles there would be with 1.60 O moles:

Finally we convert 0.18 P₃O₉ moles into grams, using its molar mass:

Explanation:

Cooking utensils, such as pots, pans and menu trays, are often made from aluminium because it is lightweight and conducts heat well, making it energy-efficient for heating and cooling. These properties also make it a preferred material for packaging.

Answer:

6.22 × 10⁻⁵

Explanation:

Step 1: Write the dissociation reaction

HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺

Step 2: Calculate the concentration of H⁺

The pH of the solution is 2.78.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M

Step 3: Calculate the molar concentration of the benzoic acid

We will use the following expression.

Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution

Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M

Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid

We will use the following expression.

Ka = [H⁺]²/Ca

Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵

Answer:

You Need To Upload The Photo Then I Will Help You :)

Explanation:

Answer:

The lithosphere solid part of the earth's crust

Solution :

It is given that :

Weight of the antacid tablet = 5.4630 g

4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.

25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.

27.7 mL of [tex]$NaOH$[/tex] solution is equivalent to [tex]$25 \ mL$[/tex] of the original stomach acid. Therefore, 13.6 mL of NaOH will take x [tex]$\frac{25\text{ mL of original stomach acid}}{\text{27.7 mL of NaOH}}$[/tex]

                                                             = 12.27 ml of the original stomach acid.

Answer:

I think it's letter B

Explanation:

Hope it helps correct if I'm wrong

it is either "aweak acid or a lousy (or very weak) acid"

Answer:

[tex]m_{H_2O}=74.8gH_2O[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible to realize there is a 1:2 mole ratio of sulfuric acid to water; thus, given the mass of the former and its molar mass (98.07 g/mol), it is possible to determine the mass of produced water as shown below:

[tex]m_{H_2O}=203.50gH_2SO_4*\frac{1molH_2SO_4}{98.07gH_2SO_4}*\frac{2molH_2O}{1molH_2SO_4} *\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=74.8gH_2O[/tex]

Regards!

Answer:

Has definite shape and volume

Solids have high melting point

Explanation:

Answer:

Concentrated sulfuric acid is used as a catalyst, and has a dual role:

-Speeds up the reaction.

-Acts as a dehydrating agent, forcing the equilibrium to the right and resulting in a greater yield of ester.

The two functions performed by the concentrated sulfuric acid in the reaction ​ is that Concentrated sulphuric acid often removes molecules of water from formic acid and it  acts as a dehydrating agent in creation of carbon monoxide.

The function of the concentrated sulfuric acid in a reaction is that Concentrated sulfuric acid can be used as a catalyst.

It help to Speeds up the reaction and can also be a dehydrating agent, by bringing in the equilibrium to the right and leading to a greater yield of ester.

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Answer:

%Cu = 66.4%

%S = 33.6%

Explanation:

Step 1: Given data

Step 2: Calculate the mass of S

The mass of CuS is equal to the sum of the masses of Cu and S.

mCuS = mCu + mS

mS = mCuS - mCu

mS = 9.56 g - 6.35 g = 3.21 g

Step 3: Calculate the percent by mass of each element

We will use the following expression.

%Element = mElement / mCompound × 100%

%Cu = mCu / mCuS × 100% = 6.35 g / 9.56 g × 100% = 66.4%

%S = mS / mCuS × 100% = 3.21 g / 9.56 g × 100% = 33.6%

Answer:

The gravitational potential energy and kinetic energy of this ball should be equal (assuming that there is no energy loss due to friction.)

Explanation:

The ball loses gravitational potential energy as it rolls down the hill. At the same time, the speed of the ball increases, such that the ball gains kinetic energy.

If there is no friction on this ball (and that the ball did not deshape,) all the gravitational potential energy that this ball lost would be converted to kinetic energy.

If the gravitational field strength [tex]g[/tex] is constant throughout, the gravitational potential energy of an object in that gravitational field would be proportional to its height.

If [tex]m[/tex] denote the mass of this ball, the gravitational potential energy ([tex]\rm GPE[/tex]) of this ball at height [tex]h[/tex] would be [tex]{\rm GPE} = (m \cdot g) \cdot h[/tex], which is proportional to [tex]h\![/tex].

The value of [tex]g[/tex] near the surface of the earth is indeed approximately constant (typically [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex].)

At halfway between the top and bottom of this hill, the height of this ball would be [tex](1/2)[/tex] of its initial value (the value when the ball was at the top of the hill.) Because the [tex]\rm GPE[/tex] of this ball is proportional to its height, at halfway down the hill, the [tex]\rm GPE\![/tex] of this ball would also be [tex](1/2)\![/tex] its initial value.

However, if there was no friction on this ball (and that the ball did not deshape,) that [tex](1/2)[/tex] of the initial [tex]\rm GPE\![/tex] of this ball was not lost. Rather, these [tex](1/2)\![/tex] of the initial [tex]\rm GPE[/tex] would have been converted to the kinetic energy ([tex]\rm KE[/tex]) of this ball.

Hence, when the ball is halfway down the hill:

[tex]\displaystyle \text{GPE halfway down the hill} = \frac{1}{2}\, \text{Initial GPE}[/tex].

[tex]\begin{aligned}& \text{KE halfway down the hill}\\ &= \text{Initial GPE} - \text{GPE halfway down the hill}\\ &= \text{Initial GPE} - \frac{1}{2}\, \text{initial GPE}\\ &= \frac{1}{2}\, \text{Initial GPE}\end{aligned}[/tex].

Therefore:

[tex]\begin{aligned}& \text{GPE halfway down the hill} \\ &= \frac{1}{2}\, \text{Initial GPE} \\ &= \text{KE halfway down the hill}\end{aligned}[/tex].

In other words, under these assumptions, when this ball is halfway down the hill, the gravitational potential energy and the kinetic energy of this ball would be equal.

Answer:

I guess this answer is c

yh

Answer:

b. (Actual yield ÷ theoretical yield) x 100

Explanation:

The percent yield will be a ratio of collected or actual yield to the theoretical yield. Like any percentage, you are determining a part of a whole. You multiply by 100 to convert your decimal answer from the division into a percentage.

Answer:

HBr has an electronegativity difference of .76 because Hydrogen's electronegativity is 2.2 and the electronegativity of Bromide is 2.96, and by the chart, it has a polar-covalent bond

Hope this helps!

Answer:

vcfg

Explanation:

The kinetic product of a reaction is the product with the high energy and favored when the temperature is low. The thermodynamic product of a reaction is the product with the low energy and favored when the temperature is high.

Regioselectivity is that part of any chemical reaction where if product is formed by more than one ways than only one way will be prefererd in this condition.

Hence correct options are high energy, low, low energy and high.

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Answer:

ANSWER is E

Explanation:

2H₃PO₄ + 3Ca(OH)₂→ 6H₂O + Ca₃(PO₄)₂

Answer:

First step is to find what the products would be. Since this seems to be a double displacement, you get: H3PO4(aq)+Ca(OH)2(aq)→ H2O(L)+Ca3(PO4)2(s)

You know the products because H has a charge of +1, so PO4 must have a charge of -3, and since OH has a charge of -1, Ca must be +2. Then make each compound have a net charge of 0. H20 is a liquid since aq solutions have liquid water. Ca3(PO4)2 is a solid, use the solubility rules.

Lastly, balance the equation: 2H3PO4(aq)+3Ca(OH)2(aq)→ 6H2O(L)+Ca3(PO4)2(s)

So the salt formed is Ca3(PO4)2

Answer:

How may we help kind sir

Explanation:

and if this was for points thanks

Answer:

a)  ≈ 4.99 / cm^3

b)   0.515 um

Explanation:

Given data:

energy = 150 Kev

dose of 10^15 cm^-2

A) find the depth of the peak of the implant profile and its value at that depth

expression for implant profile = written below in the attached file

value of the peak depth ≈ 4.99 / cm^3

B ) determine the depth at which the concentration of phosphorus is equal concentration of boron

= 0.515 um

attached below is the detailed solution to the problem

Answer:

Hyba odpowiedzi c. Hyba

Mole can be defined as the number of molar mass units of the compound in the sample. The moles of ammonium dichromate in 325 grams sample is 1.29 mol.

The molar mass can be given as the mass of each atom in the formula unit. The molar mass of ammonium chromate is given as:

[tex]\rm (NH_4)_2Cr_2O_7=2(N)+4\;\times\;2(H)+2(Cr)+7(O)[/tex]

Substituting the mass of each atom to identify the molar mass of ammonium chromate:

[tex]\rm (NH_4)_2Cr_2O_7=2(14)+4\;\times\;2(1)+2(52)+7(16)\\ (NH_4)_2Cr_2O_7=28+8+104+112\\ (NH_4)_2Cr_2O_7=252 \;g/mol[/tex]

The molar mass of ammonium dichromate s 252 g/mol. The moles of the compound in 325 g is given as:

[tex]\rm Moles=\dfrac{mass}{molar\;mass} \\\\Moles=\dfrac{325\;g}{252\;g/mol}\\\\ Moles=1.29\;mol[/tex]

The moles of ammonium dichromate in the sample are 1.29 mol. Thus, option c is correct.

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Answer:

5.68g

Explanation:

Answer:

6 volts

Explanation:

Use Ohm's Law:

V = IR

V = (2 amps)(3 ohms) = 6 volts

Answer:

[tex]\boxed {\boxed {\sf 155 \ g\ CO}}[/tex]

Explanation:

To convert from moles to mass, the molar mass must be used.

First, write the chemical formula for carbon monoxide. Since the carbon (C) comes first without a prefix, there is 1 carbon atom. The prefix mono- before oxide means 1, so there is also 1 oxygen (O) atom. The formula is CO.

Next, look up their molar masses on the Periodic Table.

Since there is 1 atom of each, the molar masses can be added.

Use this molar mass as a ratio.

[tex]\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

Multiply by the given number of moles:5.55

[tex]5.55 \ mol \ CO *\frac {28.01 \ g \ CO} {1 \ mol \ CO}[/tex]

The moles of carbon monoxide cancel.

[tex]5.55 * \frac {28.01 \ g \ CO} {1 }[/tex]

Multiply.

[tex]155.4555 \ g \ CO[/tex]

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, it is the ones place. The 4 in the tenths place tells us to leave the 5.

[tex]155 \ g\ CO[/tex]

5.55 moles of carbon monoxide is about 155 grams.

Carbon dioxide, nitrous oxide, and methane are all compounds that consist of two elements

Answer:

35.7 mL NaOH

Explanation:

M1V1 = M2V2

M1 = 0.100 M NaOH

V1 = ?

M2 = 0.102 M HI

V2 = 35.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.102 M)(35.0 mL) / (0.100 M) = 35.7 mL NaOH

when hydrogen and oxygen combine then they form water as product

also in neutralization reaction water form as product

hope it helps

Water is always a product in a type of chemical reaction called a neutralization reaction.

A reaction is a process that involves the transformation of one or more substances into one or more different substances.

Neutralization reaction is type of reaction occurs when an acid and a base react together to form a salt and water. The general equation for a neutralization reaction is:

[tex]\rm acid + base \rightarrow salt + water[/tex]

In this type of reaction, the hydrogen ions ([tex]\rm H^+[/tex]) from the acid combine with the hydroxide ions ([tex]\rm OH^-[/tex]) from the base to form water ([tex]\rm H_2O[/tex]). The remaining ions from the acid and the base combine to form a salt.

Therefore, water is always a product in a neutralization reaction, which occurs when an acid and a base react together to form a salt and water.

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