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noname [10]
3 years ago
9

for 2Na (s) +Cl2(g)->2NaCl(s) how many grams of sodium chloride will be made from 115 g of sodium metal when when reacted wit

h excess chlorine gas
Chemistry
1 answer:
alexandr1967 [171]3 years ago
5 0

 The grams  of sodium   chloride that will be made is 292.5 g


<u><em>calculation</em></u>

2Na  +Cl₂ → 2NaCl

step 1: calculate the  moles of Na

moles  =   mass/molar mass

From periodic table the  molar mass  of Na = 23 g/mol

moles = 115 g/23 g /mol =  5  moles

Step 2 : use the mole ratio to determine the  moles of NaCl

Na:NaCl   is  2:2 = 1:1  therefore the moles of NaCl  is also= 5 moles


Step  3: find the mass of NaCl

mass= moles  x molar mass

The  molar mass of NaCl = 23 + 35 .5 =58.5 g/mol

mass= 5 moles x 58.5 g/mol =292.5  g




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docker41 [41]

Answer:

The answer to your question is: b. P - Br

Explanation:

Difference of electronegativities from the periodic table. The one with the highest electronegativity will be the most polar.

a.

H =  2.2

Se = 2.55

Electronegativity = 2.55 - 2.2 = 0.35

b.

P = 2.19

Br = 2.96

Electronegativity = 2.96 - 2.19 = 0.77

c.

N = 3.04

I = 2.66

Electronegativity = 3.04 - 2.66 = 0.38

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3 years ago
16) a 110,52 g sample of mineral water is analyzed for its magnesium content. the mg2+ in the sample if first precipitated as mg
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The amount of the magnesium in the sample in ppm is 497 ppm.

<h3>What is parts per million?</h3>

The term parts per million refers to one of the manners in which concentration is expressed. It is common in oceanography and analytical chemistry.

Given that;

Mass of Mg= 0.0549 g

Mass of the sample = 110.52 g

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= 497 ppm

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1 year ago
The chemical reaction for the formation of syngas is: CH4 + H2O -&gt; CO + 3 H2 What is the rate for the formation of hydrogen,
grin007 [14]

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

CH_4+H_2O\rightarrow CO+3H_2

The expression for rate of reaction :

\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}

\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}

\text{Rate of formation of }CO=+\frac{d[CO]}{dt}

\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}

The rate of reaction expression is:

\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}

As we are given that:

+\frac{d[CO]}{dt}=0.35M/s

Now we to determine the rate for the formation of hydrogen.

+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}

+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s

\frac{d[H_2]}{dt}=3\times 0.35M/s

\frac{d[H_2]}{dt}=1.05M/s

Thus, the rate for the formation of hydrogen is, 1.05 M/s

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