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2 years ago

What is the molarity of 62grams of ammonium in 5liters of water?

Chemistry

1 answer:

Viefleur [7K]2 years ago

7 0

Answer:

Explanation:

First we need to find how many moles of ammonium weigh 62 grams.

Molar mass of NH4 = (14.0)+(4*1.0) grams

or 18.0 grams/mole

62 (g)/18(g/mole) = 3.444... moles of NH4

If it is dissolved in 5 litres of water, the concentration will be 3.444moles/5L

or 0.6888 M.

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What mass of water is produced when 425 g of octane is burned?

Vitek1552 [10]

659. 48 g of water is produced

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3 years ago

I don’t need the answers to this but u can if u want but I’m just asking what is this is asking for and how to solve it? I hope

strojnjashka [21]

Answer:

See explanation

Explanation:

Salts are produced from the reaction of an acid and a base. In general ...

Acid + Base => Salt + Weak Electrolyte

Acids from the 'Arrhenius Definition' contain an 'ionizable' hydrogen (-H). Such as, HCl, HBr, HI, HNO₃, HClO₄, HF, etc.

Bases from the 'Arrhenius Definition' contain an 'ionizable' hydroxide (-OH). Such as, LiOH, NaOH, KOH, CsOH, Ca(OH)₂, etc.

When the acid and base react, they proceed by what is known as a 'Double Replacement Reaction' or 'Metathesis Reaction'. In the process, the ions of the reactant compounds exchange positions such that a 'Driving Force' compound is formed on the product side. The Driving Force compound is <u>always</u> on the product side of the equation and is a compound that takes one of three forms => A precipitating salt, a compound of a weak acid or weak base (weaker than the starting acid or base) or gas decomposition product (~ vinegar + backing soda rxn => CO₂ gas).

For your problem, split the compound into cations and anions. You can usually tell which is which by using this format on formulas like those listed in your question => reading formula from left to right, place an imaginary line after the 1st metal => this metal will be the cation & the remaining formula will be the anion.

KBr => K | Br => K (Potassium) is the cation (K⁺) and Br (Bromide) is the anion (Br⁻)

KBr => K⁺ + Br⁻

Now. Apply this rule => Add 'H' to anion => HBr, then add 'OH' to cation => KOH.

HBr is the acid and KOH is the base. Therefore ...

Acid + Base => Salt + Wk Electrolyte

HBr + KOH => KBr + H₂O (note how ions of reactants exchange places to form products).

So, KBr comes from the reaction of acid HBr and base KOH.

HBr + KOH => KBr + H₂O

To determine the acid and base origins of LiCl and NaF, use the same logic. Hope this helps. Doc :-)

7 0

3 years ago

Which of the following is not considered an aspect of environmental health?

notka56 [123]

The one that isn't considered an aspect of environmental health such as smog<span />

6 0

3 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

2 years ago

A container initially holds 1.24mol of hydrogen gas and has a volume of 27.8L. Hydrogen gas was added to the container, and the

Fantom [35]

Answer:

After increasing the volume, we have 1.81 moles of hydrogen gas in the container

Explanation:

Step 1: Data given

Number of moles hydrogen gas (H2) = 1.24 moles

Volume of hydrogen gas (H2° = 27.8 L

The final volume is increas to 40.6 L

Step 2: Calculate the new number of moles

V1/n1 = V2/n2

⇒with V1 = the initial volume = 27.8 L

⇒with n1 = the initial number of moles H2 = 1.24 moles

⇒with V2 = the final volume = 40.6 L

⇒with n2 = the new number of moles = TO BE DETERMINED

27.8L / 1.24 moles = 40.6 L / n2

n2 = 40.6 / (27.8/1.24)

n2= 1.81 moles

After increasing the volume, we have 1.81 moles of hydrogen gas in the container

3 0

2 years ago