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Nitella [24]
3 years ago
11

Is Sr3(PO4)2 ionic or covalent

Chemistry
1 answer:
lana66690 [7]3 years ago
6 0
Sr3(PO4)2 is definitely Ionic
You might be interested in
What volume of 12M HCI is needed to prepare 250<br> of 0.20M HCI?
Alchen [17]

Answer: 4.2

Explanation:

M_{A}V_{A}=M_{B}V_{B}\\(12)V_{A}=(250)(0.20)\\V_{A}=\frac{(250)(0.20)}{12}=\boxed{4.2}

6 0
2 years ago
An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi
Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr  are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM →  54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3  . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3  .  11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0
3 years ago
PLEASE ANSWER I AM BEGGING
TiliK225 [7]

Taking into account the definition of dilution:

  • you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
  • If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.

<h3>Dilution</h3>

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

<h3>Volume of stock solution</h3>

In this case, you know:

  • Ci= 7 M
  • Vi= ?
  • Cf= 0.480 M
  • Vf= 0.120 L

Replacing in the definition of dilution:

7 M× Vi= 0.480 M× 0.120 L

Solving:

Vi= (0.480 M× 0.120 L)÷ 7 M

<u><em>Vi= 0.00823 L= 8.23 mL</em></u> (being 1 L= 1000 mL)

Finally, you will need 8.23 mL of the stock solution.

<h3>Concentration of the diluted solution</h3>

In this case, you know:

  • Ci= 7 M assuming the stock solution is 7.00 M HNO₃
  • Vi= 20 mL= 0.02 L
  • Cf= ?
  • Vf= 0.270 L

Replacing in the definition of dilution:

7 M× 0.02 L= Cf× 0.270 L

Solving:

(7 M× 0.02 L)÷ 0.270 L= Cf

<u><em>0.518 M= Cf</em></u>

Finally, the concentration is 0.518 M.

Learn more about dilution:

brainly.com/question/13505906

brainly.com/question/6692004

brainly.com/question/11931563

brainly.com/question/16343005

brainly.com/question/24709069

#SPJ1

5 0
1 year ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
Use the molar bond enthalpy data in the table to estimate the value of Δ∘rxn
MakcuM [25]

Answer:

ΔH°rxn = - 433.1 KJ/mol

Explanation:

  • CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)

∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state

∴ ΔH°CCl4(g) = - 138.7 KJ/mol

∴ ΔH°HCl(g) = - 92.3 KJ/mol

∴ ΔH°CH4(g) = - 74.8 KJ/mol

⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)

⇒  ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol

⇒ ΔH°rxn = - 433.1 KJ/mol

4 0
3 years ago
Read 2 more answers
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