How many grams of NH4Cl will dissolve in 50 grams of 90°C water?

Answer: jongho

Explanation:

Answer:

Choi JongHo

Explanation:

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

[tex]\displaystyle M_1V_1= M_2V_2[/tex]

Where M represents molarity and V represents volume.

Let the initial concentration and unknown volume be M₁ and V₁, respectively. Let the final concentration and required volume be M₂ and V₂, respectively. Solve for V₁:

[tex]\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}[/tex]

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
[tex]\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}[/tex]

Convert this value to mL:
[tex]\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}[/tex]

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

Answer:

  The solution contains 0.50 mol or 55 g of CaCl2.

Explanation:

Answer:

Germanium or Ge

Explanation:

It is on the perioidic table. Gallium has an atomic number of 31 and Germanium has an atomic number of 32.

Answer:

0.202M

Explanation:

you will want to use C₁V₁=C₂V₂ to solve for this.

C₁ is the initial concentration, in this problem it is .458 mol/L or M

V₁ is the initial volume, so it will be the 22.0 mL or 0.0220L

C₂ is the new concentration but here it is unknown.

V₂ is the new volume which is 50.0 mL or 0.0500L.

1. So now we plug this information in to the equation:

[tex](.458 mol/l)(0.0220L)=(C_{2} )(0.0500L)[/tex]

2. We then do algebra to get C₂ alone:

[tex]C_{2} =\frac{(.458 mol/L)(0.0220L)}{(0.0500L)} =0.20152 mol/L or M[/tex]

The liters cancel out and take note of the significant figures, there are 3 digits in all numbers in the question, so that means your answer must contain 3 sig figs. The answer 0.0202 rounded.

** instead of converting the mL to L you can also just keep them mL since they will be crossed out anyway. You will still get the correct answer.

Hope this help you! good luck :)

The work done by the gas against an external pressure during the isotherma expansion is -12 J.

The work done by the gas against an external pressure during the isotherma expansion is calculated as follows;

W(net) = ΔP x ΔV

where;

W(net) = (6 kPa - 8kPa) x (8L - 2L)

W(net) = - 12 kPa.L

W(net) = -12 J

Thus, the work done by the gas against an external pressure during the isotherma expansion is -12 J.

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Answer:

better sanitation systems

Answer:

The answer is Gain of electrons

Explanation:

Oxidation is the process of losing an electrons, while reduction is the process of gaining them.

Answer:

Gain of electrons

Answer:

About 0.0940 M.

Explanation:

Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

[tex]\displaystyle \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)} \longrightarrow \text{H$_2$O}_\text{($\ell$)} + \text{A}^-_\text{(aq)}[/tex]

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

[tex]\displaystyle \begin{aligned} 15.00\text{ mL} &\cdot \frac{0.125\text{ mol NaOH}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} \\ \\ &\cdot \frac{1\text{ mol OH}^-}{1\text{ mol NaOH}} \cdot \frac{1\text{ mol HA}}{1\text{ mol OH}^-}\\ \\ & = 0.00188\text{ mol HA}\end{aligned}[/tex]

Therefore, the molarity of the original solution was:

[tex]\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}[/tex]

In conclusion, the molarity of the unknown acid is about 0.0940 M.

Answer:

gain, or accept, a hydrogen cation

Explanation:

Hope this helps! :)

Answer:

# 5

Explanation:

The question describes silver being "poured" into a mold and cools to become a solid bar. This is the phase of liquid to solid. When a element cools down below it's freezing points to become a solid.

Liquid to Solid Definition:

Freezing, or solidification, is a phase transition in which a liquid turns into a solid when its temperature is lowered to or below its freezing point. All known liquids, except helium, freeze when the temperature is low enough.

The carbon atom is SP hybridized while the oxygen atom is SP2 hybrized.

The term resonance structures are those structures that can be used to explain the bonding in a molecule. Sometimes, a single structure is insufficient in explaining the structure of a compound.

Now looking at the two sturctures, Z is better because there are mo formal charges which make the molecle less stable unlike in X. In the CO2 molecule, the carbon atom is SP hybridized while the oxygen atom is SP2 hybrized.

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Answer: The entire container because gases will expand to fill it.

Explanation:

-361 kJ/mol heat would be released if 2 moles of nitrogen and 2 moles of oxygen form 4 moles of NO (g) at 101.3 kPa and 298 K.

Enthalpy of formation is defined as the heat change that is needed for formation of one mole of a substance from its constituent elements in pure state.

In the present case, Heat of formation of 4 moles of NO (g) at 101.3 kpa and 298 k is -90.25 kJ/mol. The Negative sign indicates that, heat is evolved from the system.

Thus, for 4 moles of NO, the heat of formation will be 4 X (-90.25 kJ/mol) = -361 kJ/mol.

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Answer:

3.2 M

Explanation:

M1V1=M2V2

1.8*.445=.801

.801/.25=3.204

Molarity = M2 = 3.2 M

Answer:

H2S is a hydride of sulfur while SO2 is an oxide of sulfur.

Explanation:

(a) The number of moles of gas for first case is 0.065 mol.

(b) The number of moles of gas for second case is 0.03 mol.

(c) The number of moles of gas for third case is 0.038 mol.

The number of moles of the gas is determined by applying ideal gas equation as follows;

PV  = nRT

n = PV/RT

where;

n = (1.06 x 1.25) / (0.082057 x 250)

n = 0.065 mol.

n = (0.8 x 0.925) / (300 x 0.082057)

n = 0.03 mol

n = (0.75 x 0.921)/(223 x 0.082057)

n = 0.038 mol

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Answer:

addition

Explanation:

I really don't know why

Answer:

Cooperation

Explanation:

Those animals are understanding the needs of each other.

Answer:

Oxidized, and reduced

Explanation:

The acronym LEO, and GER may be helpful, or think of a specific example (such as a copper 2+ ion being reduced into copper).
Loose
Electron
Oxidization

Gain
Electron
Reduction

from this, we know that a molecule which looses electrons is being oxidized, and a molecule that gains electrons is being reduced.  

Answer:

Weak acids and bases are less than 100% ionized in aqueous solution

Explanation:

hope this helps

pls mark brainliest

Answer:

having the same number of valence electrons

Explanation:

Answer:

the troposphere because we are living in atmosphere when the atmosphere next troposphere the sudden flashes of light and discharge resulting atmosphere condition sound waves of trambeling

thank you

Answer:

D

explaination:

the "prop" stands for three carbons , and this is an alkane so that means it only has single bonds between carbon and hydrogen , hence the name propane

Answer:

naclo4+koh naoh+kclo4

Explanation:

if you know your cemestry

Answer:

it's A

Explanation:

Answer:D)

Explanation: the distance traveled over time

A gas is contained in a tick-walled balloon. When the pressure changes from 2. 95atm to 2.57 atm. p.

Pressure is a force perpendicular to the unit area on which it is applied.

Given that, P1, the initial pressure is 2.95 atm

The initial volume, V1 = 7.456 L

The final volume, V2 = 4.782 L

The initial temperature, T1 =  379k

The final temperature, T2 = 212k

The equation will be

[tex]\dfrac{P_1V_1}{T_1 } =\dfrac{P_2V_2}{T_2}[/tex]

Putting the values in the equation

[tex]\dfrac{2.95 \times 7.456 L }{379k } =\dfrac{P_2\times 4.782 L}{212k}\\\\\\P_2 = \dfrac{2.95 \times 7.456 L \times 212k }{379k \times 4.782 L} = 2.57\;atm.[/tex]

Thus, the pressure changes from 2. 95atm to 2.57 atm p.

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