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nalin [4]
3 years ago
9

____B₂Br₆ + ___HNO₃ ---->___B(NO₃)₃ + ____HBr balance them thanks!!!

Chemistry
1 answer:
natita [175]3 years ago
6 0
1, 6, 2, 6
In the order you wrote them in
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How did I get the sand out the mixture ?
svetoff [14.1K]

Answer:

They are called homogenous mixtures. If you put sand into a glass of water, it is considered to be a mixture. You can always tell a mixture, because each of the substances can be separated from the group in different physical ways. You can always get the sand out of the water by filtering the water away.

4 0
3 years ago
Read 2 more answers
A 10.50 gram sample of a compound is decomposed to yield 3.40 g Na, 2.37 g S, and 4.73 g O. What is the mass percentage of each
9966 [12]

Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

= 32.4%

%S = 2.37/10.5 × 100%

= 22.6%

% O= 4.73/10.5 × 100%

= 45.0%xplanation:

6 0
2 years ago
In need of help very soon please.
NemiM [27]

Answer:

hello, i hope this helps.

Explanation:

1 - group

2 - period

3 - periodic table

4 - family

5 - octet rule

6 - valence electrons

6 0
2 years ago
The salts NaCl and CaCl
Irina18 [472]
The answer is 8 :)
All nobel gases have 8 outer electrons.
6 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
3 years ago
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