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RUDIKE [14]
3 years ago
13

The ion formed by an atom of tin when it loses four electrons is called what?

Chemistry
1 answer:
MatroZZZ [7]3 years ago
6 0

free radical , prositron or proton .

You might be interested in
What mass of HgO is required to produce 0.692 mol of O2?<br><br>2HgO(s) -&gt; 2Hg(l) + O2(g)
Vika [28.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

Explanation:

Given:

no of moles of the oxygen gas = 0.692

Also given:

2 HgO  → 2 Hg + O_{2}

where,

HgO represents mercuric oxide

Hg represents mercury

O_{2} represents oxygen

To calculate:

Molar mass of HgO:

Molar mass of HgO = 216 grams

molar mass of mercury (Hg) = 200 grams

molar mass of oxygen (O) =16 grams

HgO = 200 +16 = 216 grams

We know;

       2×216 grams of HgO   →  1 mole of oxygen molecule

             ?                              →  0.692 moles of oxygen molecule

       

          = \frac{2*216*0.692}{1}

      = 298.944 grams of HgO

<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>

<u />

7 0
3 years ago
Read 2 more answers
The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro
Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

6 0
3 years ago
Calculate the mass in 4.05*10^22 molecules of calcium phosphate
Marizza181 [45]

Answer:

m = 20.9 g.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:

m=4.05x10^{22}molecules*\frac{1mol}{6.022x10^{23}}*\frac{310.18g}{1mol}\\\\m=20.9g

Regards!

8 0
3 years ago
Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?
lina2011 [118]

Answer: k_a for C_6H_5NH_3^+ at 25°C is 0.588\times 10^{-4}

Explanation: We are given k_b of C_6H_5NH_2 at 25°C which is 1.7\times 10^{-9}

To calculate the k_a of C_6H_5NH_3^+, we use the formula:

k_w=k_a\times k_b

k_w\text{ at }25^o=1\times 10^{-14}

Putting values in above equation, we get:

1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}

7 0
3 years ago
A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.
lord [1]

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g

The produced energy will be:

=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K

=450.35\times 3.2

=1441.12 \ J

The reaction will be:

⇒  HCl+NaOH \rightarrow NaCl+H_{2}O

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.  

Therefore, the moles generated by NaCl will indeed be:

=  (\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}

=  0.0318\times 0.500

=  0.0159 \ mole  \ of \ NaCl

Now,

=  \frac{1441.12 \ J}{0.0159 \ moles \ NaCl}

=  906364.7

=  90.6 \ KJ/mol \ NaCl

7 0
3 years ago
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