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rodikova [14]
3 years ago
12

How many grams of molecular chlorine will be required to completely react with 0.0223 moles of sodium iodide according to the fo

llowing reaction?
2Nal + Cl2 ---> 2NaCl + I2

A. 1.57 104 grams
B. 3.16 grams
C. 0.0112 grams
D. 0.791 grams​
Chemistry
1 answer:
Alecsey [184]3 years ago
8 0

Answer: a

aa

Explanation: a

aaa

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What kinds of things could a magnet make another object or magnet do without touching it?​
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Answer:

Magnets can be used to make other magnets and things made of iron move without being touched. Something that has been electrically charged can make other things move without touching them.

Explanation:

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Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base rea
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Answer: Yes, HCl is a strong acid.

acid = HCl , conjugate base = Cl^- , base = H_2O, conjugate acid = H_3O^+

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes HCl is a strong acid as it completely dissociates in water to give H^+ ions.

HCl\rightarrow H^++Cl^-

For the given chemical equation:

HCl+H_2O\rightarrow H_3O^-+Cl^-

Here, HCl is loosing a proton, thus it is considered as an acid and after losing a proton, it forms Cl^- which is a conjugate base.

And, H_2O is gaining a proton, thus it is considered as a base and after gaining a proton, it forms H_3O^+ which is a conjugate acid.

Thus acid =  HCl

conjugate base = Cl^-

base = H_2O

conjugate acid = H_3O^+.

8 0
3 years ago
How many grams of potassium nitrate (KNO3) would form if 2.25 liters of a 1.50 molar lead nitrate Pb(NO3)2 solution reacts with
fgiga [73]
I know that the answer is 639 g
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4 years ago
A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o
Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0
4 years ago
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3 years ago
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