A 0.0250L solution of HCl is neutralized by 0.0780L of an unknown Mg(OH)2 solution. Also 0.050L of this Mg(OH)2 solution is neutralized by 0.0300L of a 0.20 M H3(PO3) solution. What is the concentration of the HCl solution

Answer:

Explanation:

molecular weight of NO₂ is 14 + 2 x 16 = 46

So one mole of nitrogen dioxide will have mass of 46 grams .

Answer:

points

Explanation:

Answer:

Where is the diagram?

Explanation:

..........................

Answer:

See explanation below

Explanation:

First, let's write the reaction:

NH₄SH(s)   <------> H₂S(g) + NH₃(g)

The reaction is already balanced so we don't need to do anything else.

Second, let's take into account the following. The Kc expression for this reaction, only compounds in gaseous state are the only ones that contribute to the equilibrium. Solid and liquid do not contribute to the Kc expression. This is because solid and liquid have a constant concentration near to 1, so, it won't do any difference.

Knowing this, the Kc expression for this reaction is:

Kc = [H₂S] [NH₃]

Now, to calculate Kp from Kc, there's an expression that helps a lot to do this. The expression is the following:

Kp = Kc (RT)ᵃⁿ    (1)

Where:

R: universal constant of gases

T: Temperature in K

ᵃⁿ = difference of the coefficients of the reaction.

This expression comes from the fact that Kp is an expression that instead of working with concentrations, it works with pressure.

If we use the ideal gas equation we have:

PV = nRT

Solving for P:

P = nRT/V        and C = n/V   so

P = CRT

If we now replace this, in the Kp expression of equilibrium we have:

Kp = pH₂S * pNH₃

Kp = ([H₂S]RT)¹ ([NH₃]RT)¹

Kp = (RT)¹⁺¹ ([H₂S] [NH₃])

Kp = (RT)²Kc

So finally the expression for Kp would be:

Hope this helps

Answer:

thermal conductors

steel

polystyrene

thermal insulator

between things of the same temp?

sorry to ask but if its ok with you, may i get brainly, i need to rank up all i need is two more, if not its fine. thank you and yours truly golden

Answer: 1.2 x 10^-17 microounces

Explanation:

Ounce = 28.5G microounce = 28.5*10^-6g

3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17

Answer:

17.5 L

Explanation:

Step 1: Given data

Step 2: Calculate the final volume of the nitrogen

We have a gas that undergoes a change at a constant temperature. If we assume an ideal behavior, we can calculate the final volume of the nitrogen using  Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1 atm × 14.0 L/0.8 atm = 17.5 L

Answer:

Different objects appear to be different colors, because we see color by light bouncing off an object and reflecting or transmitting into our eye. The amount of light that reflects or transmits into our eye depend on the color that we will see.

Answer:

0.0847M is molarity of sodium hydrogen citrate in the solution

Explanation:

The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.

Moles Na2C6H6O7:

Molar Mass:

2Na: 2*22.99g/mol: 45.98g/mol

6C: 6*12.01g/mol: 72.01g/mol

6H: 6*1.008g/mol: 6.048g/mol

7O: 7*16g/mol: 112g/mol

45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol

Moles of 2g:

2g * (1mol / 236.038g) = 8.473x10⁻³ moles

Liters solution:

100mL * (1L / 1000mL) = 0.100L

Molarity:

8.473x10⁻³ moles / 0.100L =

Answer:

0.085M

Explanation:

Molecular weight of sodium hydrogen citrate=236.09 g

236.09 g of Na2C6H6O7 in 1000ml= 1 M

2%= 2 g in 1 litre (1000ml)

20/236.09=0.0847 M=0.085 M

Answer:

1.6 °C.kg/mol

Explanation:

Step 1: Calculate the molality of urea

We will use the following expression.

m = mass(urea) / molar mass(urea) × kg solvent

m = 24.6 g / 60.06 g/mol × 0.650 kg

m = 0.630 mol/kg

Step 2: Calculate the boiling point elevation of X

The boiling point elevation is a colligative property that can be calculated using the following expression.

ΔTb = 124.3 °C - 123.3 °C = 1.0 °C

Step 3: Calculate the boiling point elevation constant

We will use the following expression.

ΔTb = Kb × m

Kb = ΔTb/m

Kb = 1.0 °C/(0.630 mol/kg) = 1.6 °C.kg/mol

Answer:

1.2 cm

Explanation:

The reason how I got my answer is by dividing 150 by 125 and that gave me 1.2 cm.

Answer: The empirical formula is [tex]C_3H_3O[/tex].

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2[/tex] = 12.24 g

Mass of [tex]H_2O[/tex] = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, =[tex]\frac{12}{44}\times 12.24=3.338g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, =[tex]\frac{2}{18}\times 2.505=0.278g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles[/tex]

Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles[/tex]

Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =[tex]\frac{0.278}{0.104}=3[/tex]

For H =[tex]\frac{0.278}{0.104}=3[/tex]

For O =[tex]\frac{0.104}{0.104}=1[/tex]

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is [tex]C_3H_3O[/tex].

Answer:

C. P = nRT

Explanation:

PV = nRT, where n is a number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K.

Hope this helps :)

Answer:T =PV/nR

Explanation:

Answer:

3.2 × 10³ J

Explanation:

Step 1: Given data

Step 2: Calculate the temperature change (ΔT)

ΔT = 1064 °C - 75 °C = 989 °C

Step 3: Calculate the energy required (Q)

We will use the following expression.

Q = Cp × m × ΔT

Q = 0.129 J/g.°C × 25 g × 989 °C

Q = 3.2 × 10³ J

Answer:

O Each orbit has a specific energy level.

Explanation:

Neils Bohr put forward his own model of the atom based on the quantum mechanics originally developed by Planck.

He assumed the Rutherford's model and suggested that the extranuclear part consists of electrons in specific spherical orbits around the nucleus.

The orbits/energy level are the permissible through which the electrons can move through.

His concept is based the concept that the electron can move round the nucleus in certain permissible orbitss

Answer:

1. 0.74mol

2. 0.42mol

3. 2.125mol

4. 0.301mol

5. 4.52 × 10^23 particles

Explanation:

Number of moles (n) in a substance can be found using the formula:

mole (n) = mass/molar mass

Using this formula, the following moles are calculated:

1. Molar of Na = 23g/mol

mole = 17/23

mole = 0.74mol

2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)

= 46 + 32 + 64

= 142g/mol

Mole = 60/142

mole = 0.42mol

3. Molar mass of CO2 = 12 + 16(2)

= 12 + 32

= 44g/mol

mole = 93.5/44

mole = 2.125mol

4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)

= 23 + 14 + 48

= 85g/mol

mole = 25.6/85

mole = 0.301mol

5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;

0.75mol × 6.02 × 10^23

= 4.515 × 10^23

= 4.52 × 10^23 particles

Answer:

Unlike a permanent magnet, an electromagnet can be turned on and off using electrical current. Many variables affect the strength of this electromagnet, and there are some variables that do not affect the strength. ... Making the nail longer will not make the magnet stronger, unless you also add more turns to the coil.

electromagnet can be turned on and off using electrical current. Many variables affect the strength of this electromagnet, and there are some variables that do not affect the strength

Answer:

[tex]d=7.77\times 10^{13}\ m[/tex]

Explanation:

We know that the speed of light is equal to [tex]3\times 10^8\ m/s[/tex]

We need to find how far the light travel in 3 days.

Speed of an object is equal to distance covered divided by time taken.

Also, 1 day = 86400 s

3 days = 259200 s

So,

[tex]d=vt[/tex]

[tex]d=3\times 10^8\times 259200\\\\d=7.77\times 10^{13}\ m[/tex]

So, the light will travel [tex]7.77\times 10^{13}\ m[/tex] in 3 days.

An SDOF system is one whose motion is governed by a single, second-order differential equation. Only two variables, position and velocity are needed to describe the trajectory of the system. Many structures can be idealized as single degree-of-freedom systems.

The LDAP Data Interchange Format is a standard plain text data interchange format for representing LDAP directory content and update requests. LDIF conveys directory content as a set of records, one record for each object.

its dry i think

Explanation:

it might be dry

Answer:

Mass =  76986  g

Explanation:

Given data:

Dimensions of tank = 126 cm× 47 cm× 13 cm

Mass of water required to filled = ?

Solution:

First of all we will calculate the volume of tank which is equal to the volume of water required to fill it.

Volume = length ×height ×width

Volume = 126 cm × 13 cm× 47 cm

Volume = 76986 cm³

Mass of water:

Mass = density × volume

density of water is 1 g/cm³

Mass = 1 g/cm³× 76986 cm³

Mass =  76986  g

Answer:

hello your question is incomplete attached below is the complete question

answer :

Absence of c = c = C-H of the peak and the presence of -O-H and C-O peak shows that there is presence of Alcohols as products and it also shows that there is no starting material left

Explanation:

To provide that there is no starting material left and products are alcohols using the IR spectrum we will apply hydroboration oxidation

Absence of c = c = C-H of the peak and the presence of -O-H and C-O peak shows that there is presence of Alcohols as products and it also shows that there is no starting material left

attached below is the detailed solution

The empirical formula : C. C₁₆H₁₅N₂

Given

12.5 g of aniline

7.1 grams of H2O

1.48 grams of N2

Required

The empirical formula

Solution

Reaction :

mass H in H₂O :

= 2.1/18 x 7.1 g

= 0.79

mass N = 1.48

mass C :

= 12.5 g-(mass H+mass N)

= 12.5 - (0.79+1.48)

= 10.23

Mol ratio C : H : N =

= 10.23/12 : 0.79/1 : 1.48/14

= 0.853 : 0.79 : 0.106

= 8 : 7.5 : 1

= 16 : 15 : 2

Answer: electrons

Explanation:

Answer: It is called an electron because it is attracted to the positively charged particales called protons in the center of the atom.

Explanation:

Answer:

See explanation below

Explanation:

You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.

Now let's analyze what is happening in the reaction so we can predict the final product.

We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.

Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.

Hope this helps

Answer:

42.658

Explanation:

1) Avogadro's number is: N(A)=6.02*10²³

2) M of (Mo₂(SO₄)₃)=96*2+96*3=480 (gr/mol);

3) if N(A) ~ 1 mol of (Mo₂(SO₄)₃), tnen 5.35*10²² ~

(5.35*10²²)/(6.02*10²³)≈0.0888 mol of (Mo₂(SO₄)₃);

4) m(Mo₂(SO₄)₃)=M(Mo₂(SO₄)₃)*0.0888;

m((Mo₂(SO₄)₃)=480*0.0888=42.658 (gr)

Answer:

25.33 grams of water are produced.

Explanation:

15 g H / 1.00784* = 14.8833 moles H

7.5 g O / 15.999** = 0.468779 moles O

Ratio of Hydrogen to Oxygen in Water: 2 : 1

Oxygen: 0.468779 moles

Hydrogen (calculated based on oxygen): 0.468779 * 2 = 0.937558 moles

Water: 0.468779 + 0.937558 = 1.406337 moles

Grams of water produced:

1.406337 mol * (water molar mass)

= 1.406337 mol * [2(1.00784) + 1(15.999)]

= 1.406337 mol * (18.01468***)

= 25.3347110272 g water

Notes:

*molar mass of hydrogen

**molar mass of oxygen

***molar mass of water is actually 18.01528, so the calculation of 18.01468 is close enough

-------------------------------- Summary --------------------------------

Convert hydrogen and oxygen to moles.

Determine number of moles of water using: 2 mole hydrogen + 1 mole oxygen = 1 mole water

Convert water moles to grams.

Answer:

b

Explanation:

it is not c because water expands when it freezes.

Answer:

the answer is asteroids!

Explanation:

Answer:

astroid

Explanation:

becaus