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Bad White [126]
3 years ago
13

Does anyone have any tips on studying the 17 bones for the test on Monday?

Chemistry
1 answer:
LUCKY_DIMON [66]3 years ago
7 0
Whenever I have tests like these, I make a quizlet with each part zoomed in and then answer it that way.
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What is the name for NiCO3 • 4H2O
lianna [129]
<span>Nickel carbonate tetrahydrate</span>
6 0
3 years ago
Read 2 more answers
How many moles is in 2.52*10^24 molecules of water?
nataly862011 [7]
Hi friend
--------------
Your answer
-------------------

Water = H2O

Number of molecules in one mole of water = 6.022 × 10²³ [Avogadro's constant]

Given number of molecules = 2.52 × 10²³

So,
------

Number of moles =
\frac{2.52 \times 10 {}^{24} }{6.022 \times 10 {}^{23} }  \\  \\  = 4.184 \: (approximately)

HOPE IT HELPS
3 0
3 years ago
Given: N2H4(l) + O2(g) LaTeX: \longrightarrow⟶ N2(g) + 2H2O(g) ΔH°1 = –543 kJ·mol–1 2H2(g) + O2(g) LaTeX: \longrightarrow⟶ 2H2O(
kakasveta [241]

Answer:

Explanation:

Given reaction

N₂H₄ + O₂ ⇒ N₂ + 2H₂O   ΔH₁ = -543 KJ ---------- ( 1 )

2H₂ + O₂ ⇒ 2H₂O              ΔH₂ = -484 KJ ---------- ( 2 )

N₂ + 3 H₂ ⇒ 2NH₃              ΔH₃ = -92 KJ  -----------( 3 )

( 1 ) -  ( 2 ) +( 3 )

N₂H₄ + O₂ - 2H₂ - O₂ +N₂ + 3 H₂ ⇒ N₂ + 2H₂O - 2H₂O +2NH₃  

                                                         ΔH =       -543 + 484 -92 = -151 KJ

     N₂H₄ + H₂ ⇒ 2NH₃     ΔH  = -151 KJ .

2NH₃ ⇒ N₂H₄ + H₂     ΔH  = + 151 KJ

3 0
3 years ago
Which group and period does the electron figuration represent 1s22s22p63s24s23d8
taurus [48]

<em>grou</em><em>p</em><em> </em><em>4</em><em> </em><em>,</em><em> period</em><em> </em><em>4</em>

Explanation:

from the electron configuration,

the element is Titanium

and it's has 22 electrons on it's shells , so wen you place it on the periodic table, you will see that it's in group 4 period 4

5 0
3 years ago
What is the angular momentum of last electron of chromium?
damaskus [11]

Answer:

expected configuration for Cr is

1s2 2s2 2p6 3s2 3p6 4s2 3d4

but in real, it is

1s2 2s2 2p6 3s2 3p6 4s1 3d5

electron from 4s orbital jumps to 3d orbital to get stable configuration.

so the last electron comes in 3d orbital as filling of 3d takes place after filling of 4s orbital.

Hence,quantum numbers for last electron in Cr is :-

n = 3

l = 2

m = +2

s = +1/2

Explanation:

5 0
4 years ago
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