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2 years ago

In the reaction Mg(s) + Br2() → MgBrz(s), what should be the coefficient for

Chemistry

1 answer:

lbvjy [14]2 years ago

6 0

Answer:

Explanation:

Chemical equation:

Mg + Br₂ → MgBr₂

Balanced chemical equation:

Mg + Br₂ → MgBr₂

Left hand side Right hand side

Mg = 1 Mg = 1

Br = 2 Br = 2

There are one Mg and 2 Br atoms on both side of equation thus this equation is balanced. Thus coefficient with bromine is 1.

Mg is present in group 2 it loses two valance electrons and get +2 oxidation state. Bromine is present in group 17 have 7 valence electrons. Thus one bromine atom need only one electron to complete the octet that's why two bromine atoms will attached with one magnesium atom to cancel the charge +2 and make the compound neutral.

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evablogger [386]

Answer:

the volume of the same gas at pressure of 1.00atm =737.3ml

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2 years ago

The microscopes used today are just like the ones used by Leeuwenhoek and Hooke. True or false

Dmitry [639]

Answer:

False

Explanation:

While we do know that A. Leeuwenhoek used a simple microscope that consisted of only 1 lens, Hooke used a compound microscope. Although, after trying a compound microscope, Hooke found out that it strained his eyes and continued to use a simple microscope for his <em>Micrographia</em>.

Thus, we can say that the (compound) microscopes used today are different than the (simple) microscope used by Hooke and Leeuwenhoek.

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1 year ago

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

Leto [7]

Answer: The nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

$_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}$

For the given equation :

$^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}$

As the atomic number and mass number must be equal on both sides of the nuclear equation:

$^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}$

Thus the nuclide symbol of X is $^{231}_{90}\textrm{Th}$

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3 years ago

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Answer:

10.726g

Explanation:

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2 years ago

Read 2 more answers

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ElenaW [278]

Answer:

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2 years ago