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3 years ago

Plz help me with this:

Chemistry

1 answer:

In-s [12.5K]3 years ago

7 0

Answer:

1) H 1s^1

2) [Ar] 3d⁸ 4s²

3) [He] 2s2 2p3

4) [Kr] 4d10 5s2 5p5

5) [Ar] 4s² and [Ar] 4s²

6) [He] 2s2 2p2

7) [He] 2s² 2p⁴

8) [Ar] 3d7 4s2

9) [Kr] 4d¹⁰ 5s¹

10) [Ne] 3s² 3p⁶

Explanation:

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What is the percent composition by mass of sulfur in the compound mgso4

Margaret [11]

Answer:

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

Explanation:

Molar mass of compound = 120 g/mol

Number of sulfur atom = 1

Atomic mass of sulfur = 32 g/mol

Percentage of element in compound :

$=\frac{\text{Number of atoms}\times \text{Atomic mass}}{\text{molar mas of compound}}\times 100$

Sulfur :

$=\frac{1\times 32 g/mol}{120 g/mol}\times 100=26.7\%$

26.7% is the percent composition by mass of sulfur in a compound named magnesium sulfate.

6 0

3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

When magnesium reacts with hydrochloric acid, hydrogen gas is formed: 2HCl + Mg → H2 + MgCl2. What is the volume of hydrogen pro

Maksim231197 [3]

Answer:- C. 16.4 L

Solution:- The given balanced equation is:

$2HCl+Mg\rightarrow H_2+MgCl_2$

From this equation, there is 2:1 mol ratio between HCl and hydrogen gas. First of all we calculate the moles of hydrogen gas from given grams of HCl using stoichiometry and then the volume of hydrogen gas could be calculated using ideal gas law equation, PV = nRT.

Molar mass of HCl = 1.008 + 35.45 = 36.458 gram per mol

The calculations are shown below:

$49.0gHCl(\frac{1molHCl}{36.458gHCl})(\frac{1molH_2}{2molHCl})$

= $0.672molH_2$

Now we will use ideal gas equation to calculate the volume.

n = 0.672 mol

T = 25 + 273 = 298 K

P = 101.3 kPa = 1 atm

R = $0.0821\frac{atm.L}{mol.K}$

PV = nRT

1(V) = (0.672)(0.0821)(298)

V = 16.4 L

From calculations, 16.4 L of hydrogen gas are formed and so the correct choice is C.

7 0

3 years ago

What is the molarity of the Ca(OH)2 solution if 32.00 mL of Ca(OH)2 requires 16.08 mL of a 2.303 M solution for complete titrati

amm1812

Answer:

For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb

Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L

So now we can plug into the equation and solve:

1.338 mol/L * 0.01875 L= x mol/L * 0.02473 L

This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.

First step: setup the stoichiometry problem:

3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl

Second step: Molarity equation

Molarity= moles HCl/ L M= mol HCl/0.04027 L

For the third problem, you will just use the same equation as the first: MaVa= MbVb

Ma=0.57 M Va= x L Mb= 0.875 M Vb= 23.83 mL= 0.02383 L

0.57 M * x L= 0.875 M * 0.02383 L

With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.

0.75 M= x mol/ 0.0227 L mol NaOH = 0.75 M *0.0227 L mol NaOH= 0.017025

So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles

Molar Mass= 3.6 g/ 0.017025 mol

And with that you will find the molar mass of HX, and even determine what X is.

Explanation:

hope I helped

8 0

2 years ago

If a lab requires each a lab group (3 students) to have 25 ml of a solution and it takes 15 grams of AgNO₃ cuprous nitrate, to m

pochemuha

Answer:

0.375 grams are needed to make 25 mL solution.

Explanation:

Mass of $AgNO_3$ cuprous nitrate required to make 1 l of solution = 15 g.

1 L = 1000 mL

Mass of $AgNO_3$ cuprous nitrate required to make 1000 mL of solution = 15 g

Mass of $AgNO_3$ cuprous nitrate required to make 1 mL of solution:

$=\frac{15}{1000} g$

Mass of $AgNO_3$ cuprous nitrate required to make 25 mL of solution:

$=\frac{15}{1000} \times 25 g=0.375 g$

0.375 grams are needed to make 25 mL solution.

4 0

3 years ago