What happens when the elements in group 2 react with water?

Answers

Answer 1

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

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Related Questions

what’s the most abundant isotope of lawrencium

Answers

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you

The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .

Answers

Answer:

Explanation:

Kb of  (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N   a   is .48 M

(CH₃)₃N    +   H₂O =  (CH₃)₃NH⁺  +  OH⁻

a - x                                     x               x  

x² / (a - x )  = Kb

x is far less than a so a - x can be replaced by a .

x² / a   = Kb

x²  = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

What is the Kc equilibrium-constant expression for the following equilibrium? S8(s) + 24F2(g) 8SF6(g)

Answers

Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

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A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical formula for each compound.

a. 1.245 g Ni, 5.381 g I,
b. 2.677 g Ba, 3.115 g Br,
c. 2.128 g Be, 7.557 g S, 15.107 g

Answers

Answer:

you can see the empirical formula at the pic

The empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

What is empirical formula?

Empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a compound.

(a) 1.245 g Ni : 5.381 g I

Mole of Ni ; Mole of I = 1.245/59 : 5.381/127 = 0.02 : 0.04 = 1:2

So the formula is NiI2

(b) 2.677 g Ba : 3.115 g Br

Mole of Ba : Mole of Br = 2.677/137 : 3.115/60 = 0.019 : 0.038

= 0.02 : 0.04 = 1:2

So the formula is BaBr2

(c) 2.128 g Be : 7.557 g S

Mole of Be : Mole of S = 2.128/9 : 7.557/32 = 0.2 : 0.2 = 1:1

So the formula is BeS

Thus, empirical formula for compound (a) is NiI2, (b) is BaBr2 and (c) is BeS.

To learn more about empirical formula, refer to the link below:

https://brainly.com/question/11588623

#SPJ2

   

When does carbon dioxide absorb the most heat energy?
during freezing
during deposition
during sublimation
during condensation

Answers

During sublimation

This has been posted on here before so you could’ve searched it lol.

Best of luck :))

Answer:

during sublimation

Explanation:

just took the test

A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain

Answers

Answer:

[tex]Total = 50.6\ moles[/tex]

Explanation:

Given

[tex]Propane = C_3H_8[/tex]

Represent Carbon with C and Hydrogen with H

[tex]C = 13.8[/tex]

Required

Determine the total moles

First, we need to represent propane as a ratio

[tex]C_3H_8[/tex] implies

[tex]C:H = 3:8[/tex]

So, we're to first solve for H when [tex]C = 13.8[/tex]

Substitute 13.8 for C

[tex]13.8 : H = 3 : 8[/tex]

Convert to fraction

[tex]\frac{13.8}{H} = \frac{3}{8}[/tex]

Cross Multiply

[tex]3 * H = 13.8 * 8[/tex]

[tex]3 H = 110.4[/tex]

Solve for H

[tex]H = 110.4/3[/tex]

[tex]H = 36.8[/tex]

So, when

[tex]C = 13.8[/tex]

[tex]H = 36.8[/tex]

[tex]Total = C + H[/tex]

[tex]Total = 13.8 + 36.8[/tex]

[tex]Total = 50.6\ moles[/tex]

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