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3 years ago

Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states.

Chemistry

1 answer:

Lilit [14]3 years ago

6 0

Answer:

Check Explanation.

Explanation:

Formation reactions are chemical reactions where one mole of a compound is produced from its constituent elements in their standard states.

NaBr (s)

The Standard formation reaction is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

Using appendix C, the standard heat of formation of NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The Standard formation reaction is

S (s) + (3/2) O₂ (g) → SO₃ (g)

Using appendix C, the standard heat of formation of SO₃(g) is

ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The Standard formation reaction is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

Using appendix C, the standard heat of formation of Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

Hope this Helps!!!

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What statement best explains how life functions a unicellular organism are carried out?

s344n2d4d5 [400]

Answer:

The structures in the cell work together to perform its life functions

Explanation:

4 0

3 years ago

Read 2 more answers

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the

BARSIC [14]

Answer:

Rb+

Explanation:

Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.

Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.

Thus our calculations are:

V = 17.0 mL x 1 L / 1000 mL = 0.017 L

2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol

0.0425 mol = 4.36 g/ MW XOH

MW of XOH = (atomic weight of X + 16 + 1)

so solving the above equation we get:

0.0425 = 4.36 / (X + 17 )

0.7225 +0.0425X = 4.36

0.0425X = 4.36 -0.7225 = 3.6375

X = 3.6375/0.0425 = 85.59

The unknown alkali is Rb which has an atomic weight of 85.47 g/mol

6 0

4 years ago

calculate the ph for each case in the titration of 50.0 ml of 0.130 m hclo(aq) with 0.130 m koh(aq). use the ionization constant

-BARSIC- [3]

The ph before the addition of any Koh is 10.105.

Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.

After the addition of 50 ml KOH,

moles of KOH = 50 * 0.13 = 6.5 mmol

moles of HClO = 50 * 0.13 = 6.5 mmol

occurred hydrolysis solution,

pH = 0.5(14 + pKa + log [base conjugate])

pH = 0.5(14 + (- log (4 * 10^-8)) + log (6.5/(50 + 50)))

pH = 10.105

The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.

The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.

Learn more about concentration here:-brainly.com/question/26255204

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4 0

2 years ago

Please help me to answer this question. i have to submit tomorrow.

ratelena [41]

Answer:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

4 0

3 years ago