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lesya [120]
3 years ago
6

List the number of each type of atom on the right side of the equation pb(no3)2(aq)+2nacl(aq)→pbcl2(s)+2nano3(aq)

Chemistry
2 answers:
Lena [83]3 years ago
6 0
 The  number   of each  type  of atom  on the  right  side  of the equation

that is

Pb(No3)2(aq)  + 2 NaCl(aq)  =     PbCl2(s)   +2NaNo3(aq)


are    1  atom  of  Pb
          2 atoms of   N
          6 atoms  of    O
          2  atoms  of  Na
           2  atoms  of Cl


Explanation
In  Pb(NO3)2  there  is  1 x1=1  atoms  of  Pb,   1x2 =2  atoms   of N  and   3x2 =6  atoms of  O

In   NaCl  there is    2 x1= 2  atoms of Na  and  2 x1=2  atoms  of  Cl
Savatey [412]3 years ago
4 0
<span>Answer:

Pb: 1 atom

Cl: 2 atoms

Na: 2 atoms

N: 2 atoms

O: 6 atoms


Explanation:


1) RighT side of the equation: PbCl₂(s)+2NaNO₃(aq)
</span><span />

<span>2) The coefficients and subscripts tells the number of atoms in each chemical formula by this convention:
</span><span />

<span>i) The coefficient before the compound (chemical formula) applies to all the elements to the right (in the very chemical formula).


</span><span>ii) The subscripts to the right of the element apply only to the element to which it is subscripted.</span>
<span /><span>
ii) If there are parenthesis the subscrpts apply to all the elements inside the parenthesis.
</span><span />

<span>3) With that, let's see each compound (chemical formula):
</span><span />

<span>i) Pb Cl₂ (s): there are two kind of atoms, Pb and Cl.
</span><span />

<span>Pb does not have any subscriPT, which is interpreted as subscript 1, so that is 1 atom of Pb.
</span><span />

<span>Cl₂ means 2 atoms of chlorine
</span><span />

<span>ii) 2NaNO₃:
</span><span />

<span>Na: 2 atoms (because the coefficient)

N: 2 atoms (because the coefficient)

</span><span>O: 6 atoms ( 2 x 3)
</span><span>
</span>
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Vitek1552 [10]

Answer: In simplest case mass of reactants is same as mass of products.

Without thinking this question deeper, mass of ZnCl2 would be 49, but..

Explanation: Reaction should be  Zn + 2 HCl ⇒ ZnCl2 + H2

Amount of zinc is  5 g / 65,38 g/mol = 0,076476 mol and amount

of Hydrogen Chloride is 50 g / 36.458 g/mol = 1,371 mol.

Althought HCl is needed 0.152 moles, zinc is an limiting reactant.

So it is possible to produce only 0.076476 mol Hydrogen and its mass

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4 0
2 years ago
After 17.1 thousand years, what percentage of the original carbon-14 would be left in an organism’s remains? After 17.1 thousand
hram777 [196]

Answer:

12.5%

Explanation:

Initial percentage of carbon 14 = 100%

Final percentage = ?

Time passed = 17.1 * 1000 years = 17100 years

Half life of carbon 14 = 5,730 years.

So how many Half lives are in 17100 years?

Number of Half lives = Time passed /  Half life = 3.18 ≈ 3

First Half life;

100% --> 50%

Second Half life;

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Third Half life ;

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2 years ago
How many moles of H20 are needed to produce 1 mole of O2 in the following equation?
Alex787 [66]
2 moles is the answer
6 0
3 years ago
05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises
OLga [1]

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

6 0
3 years ago
What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-
Sergio [31]

Answer:

We need 4.28 grams of sodium formate

Explanation:

<u>Step 1:</u> Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

<u>Step 2:</u> Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

<u>Step 3:</u> Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

<u>Step 4:</u> Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

4 0
3 years ago
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