What is the molarity of a solution that contains 25.3 g of NaCl in 111 mL of solution?

-14.6 M
-3.90 M
-0.25 M
-5.52 M

Answer:

0.05M is the concentration of the phosphate buffer

Explanation:

A buffer is defined as the aqueous mixture of a weak acid (KH₂PO₄) and its conjugate base (K₂HPO₄). The concentration of this buffer is the sum of the concentrations of the weak acid and the conjugate base, that is:

Concentration of phosphate buffer = 0.02M + 0.03M

=

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.

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

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.

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:

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:

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:

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: 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:

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:

the answer is asteroids!

Explanation:

Answer:

astroid

Explanation:

becaus

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:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density [tex]\rho_{avg}[/tex]; we have:

[tex]\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex]

The average atomic weight is:

[tex]A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex]

So; in terms of vanadium, the Concentration of iron is:

[tex]C_{Fe} = 100 - C_v[/tex]

From a unit cell volume [tex]V_c[/tex]

[tex]V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}[/tex]

where;

[tex]N_A[/tex] = number of Avogadro constant.

SO; replacing [tex]V_c[/tex] with [tex]a^3[/tex] ; [tex]\rho_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex] ; [tex]A_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex] and

[tex]C_{Fe}[/tex] with [tex]100-C_v[/tex]

Then:

[tex]a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }[/tex]

[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }[/tex]

[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }[/tex]

Replacing the values; we have:

[tex](0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }[/tex]

[tex]2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }[/tex]

[tex]2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})[/tex]

[tex]\mathbf{C_v = 9.1 \ wt\%}[/tex]

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:

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

its dry i think

Explanation:

it might be dry

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: 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:

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:

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:

Explanation:

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

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

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:

b

Explanation:

it is not c because water expands when it freezes.

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:

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:

Atoms with a positive charge will be attracted to negatively charged atoms to form a molecule. This bonding between atoms is the key to how molecules interact with each other. ... Other molecules interact with water, combining with the atoms to create new soluble liquids.

Answer:

points

Explanation:

Answer:

Where is the diagram?

Explanation:

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