Ask question

Ask question

All categories

3 years ago

15

An unknown compound has the following chemical formula: where stands for a whole number. Measurements also show that a certain s

ample of the unknown compound contains of magnesium and of chlorine. Write the complete chemical formula for the unknown compound.

1 answer:

nirvana33 [79]3 years ago

8 0

Answer: This is only a guess: The "following chemical formula: " doesn't appear on my screen. But since you mentioned Mg and Cl, I would guess MgCl2

Explanation: Mr loses 2 electrons while each Cl gains one each: Two chlorines for every 1 magnesium.

You might be interested in

What are the products?

LuckyWell [14K]

I think it would be option 4?

7 0

3 years ago

Read 2 more answers

Which is it , if you are right I will give brainliest

Elan Coil [88]

Answer:

a

Explanation:

6 0

3 years ago

Read 2 more answers

Explain how you would make 450 mL of a .250M NaOH solution?

Neporo4naja [7]

Answer:

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

Explanation:

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of NaOH = x

Molar mass of NaOH = 40 g/mol

Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)

Molarity of the solution of NaOH = 0.250 M

$Molarity=\frac{1.248 g}{26 g/mol\times 0.9102254 L}=0.528 mol/L$

$0.250 M=\frac{x}{40 g/mol\times 0.450 L}$

Solving for x:

x = 4.5 g

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

4 0

4 years ago

Determine the percent composition of sodium in sodium carbonate

crimeas [40]

11×2=22
(11×2)+(6)+(8×3)=52
22/52=0.4230
0.4230×100=42.3%

3 0

3 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

6 0

3 years ago