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3 years ago

One of the scientists interviewed for the article said, "Almost certainly this

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

The scientist is referring to the unusually slow growth rate of carbon dioxide occurring recently. As that slow growth rate ends, the rate of increase of carbon dioxide in the atmosphere should go up soon.

Explanation:This is the exact answer so switch it up.

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The density of a silver is always ____

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The density of silver is 10.49 g/cm^3

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Hi how has your day been

Lady bird [3.3K]

Answer:

good man thanks

Explanation:

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Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

choli [55]

Explanation:

a) $H_2SO_4$

oxidation state of H = +1

Oxidation state of O = -2

Let the oxidation no. of S be x.

$2\times1+ x+2\times -4 =0$

$2 +x-8 = 0\\x=+6$

Oxidation state of S in $H_2SO_4$ = +6

b) $Ca(OH)_2$

oxidation state of H = +1

Oxidation state of O = -2

Let the oxidation no. of Ca be x.

$x+2\times-2+2\times+1 =0\\x=4 -2\\x=+2$

Oxidation state of Ca in $Ca(OH)_2$ = +2

c) BrOH

oxidation state of H = +1

Oxidation state of O = -2

Let the oxidation no. of Br be x.

$x+1\times-2+1\times+1 =+1$

Oxidation no. of Br in BrOH is +1

d) $ClNO_2$

oxidation state of Cl = -1

Oxidation state of O = -2

Let the oxidation no. of N be x.

$-1+x+2\times-2\\x = 4+1 =+5$

Oxidation state of N in $ClNO_2$ = +3

e) $TiCl_4$

oxidation state of Cl = -1

Let the oxidation no. of Ti be x.

$x+4\times -1 = 0\\x=+4$

Oxidation state of Ti in $TiCl_4$ = +4

f) NaH

Na is more electropositive than H.

Therefore, oxidation state of Na = +1

Oxidation state of H = -1

6 0

3 years ago

What amount of oxygen gas in moles contains 1.8x10^22 molecules?

exis [7]

Answer:

<h2>0.03 moles </h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

We have

$n = \frac{1.8 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.02990...$

We have the final answer as

<h3>0.03 moles </h3>

Hope this helps you

7 0

3 years ago

Refer to the electron distribution to answer 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹. What would be the period or series number of

Salsk061 [2.6K]

Answer:

C.5

Explanation:

A number of electrons present in valence shell of penultimate Shell represents the group of elements.

For s block elements: no.of group=number of valence shell electron.

p block elements: no. of group= 2 + 10 + number of valence electrons.

d block elements: no. of group= number(n-1)d electrons + number of electrons in nth shell.

Here, the differential electron is in p orbital hence, it belongs to p block

No. of group= 2+ 10 + 3=15 i.e 15th group or VA group.

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