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2 years ago

5

What is the formula for silicon tetrabromide?

2 answers:

Kaylis [27]2 years ago

7 0

Answer:

SiBr4

<em>hope this helps!</em>

bogdanovich [222]2 years ago

5 0

Answer:

SiBr4

Explanation:

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Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

Thus, for all the given combinations, we obtain:

- Y⁻

$X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3$

- Y²⁻

$X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3$

- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

Best regards!

8 0

2 years ago

If 25 mL of a HCl solution of unknown concentration was neutralized with 10 mL of a 0.30 M NaOH solution, what was the original

QveST [7]

Answer:

0.12 M

Explanation:

Step 1: Write the balanced equation

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of NaOH

10 mL of a 0.30 M NaOH solution react.

$0.010L \times \frac{0.30mol}{L} = 3.0 \times 10^{-3} mol$

Step 3: Calculate the reacting moles of HCl

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 1/1 × 3.0 × 10⁻³ mol = 3.0 × 10⁻³ mol.

Step 4: Calculate the concentration of HCl

3.0 × 10⁻³ mol of HCl are in 25 mL of solution.

$M = \frac{3.0 \times 10^{-3} mol}{0.025L} = 0.12 M$

5 0

2 years ago

Are pizza boxes recyclable?

shtirl [24]

Yes they are because they are cardboard and people can make things out of cardboard

4 0

2 years ago

Name the only satellite which is known to have an atmosphere in the planet it orbits

Maurinko [17]

Answer:

Titan which orbits Saturn

Explanation:

6 0

2 years ago

A 50 gram sample of a radioisotope undergoes 2 half-lives. How many grams

Veseljchak [2.6K]

Answer:

At the end of second half life 12.5 g will left

Explanation:

Given data:

Total Mass = 50 g

Half lives = 2

Mass remain at the end = ?

Solution:

At time zero = 50 g

At 1st half life = 50 g /2 = 25 g

At second half life = 25 g/2 = 12.5 g

So at the end of second half life 12.5 g will left.

3 0

2 years ago