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3 years ago

True or False - Elemental gases are relatively insoluble in water and should be given the state (g).

Chemistry

1 answer:

neonofarm [45]3 years ago

3 0

Answer:

True

Explanation:

I assume this is refering to gases such as O2, H2, I2, F2, ect. And possibly the nobel gases which are completly insoluable. As gases they should be given the state (g).

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How many milliliters of an aqueous solution of 0.227 M sodium carbonate is needed to obtain 3.55 grams of the salt?

Marianna [84]

Answer:

147.6 mL .

Explanation:

sodium carbonate , Na₂CO₃

Molecular weight = 106

3.55 gm of sodium carbonate = 3.55 / 106

= .0335 moles

Let the volume of litre required = V

V litre of .227 M solution will contain

V x .227 moles of sodium carbonate . So

V x .227 = .0335

V = .1476 L

= 147.6 mL .

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Its D as A, B and C are physical properties of water

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Please use the values in the resources listed below instead of the textbook values. Under certain conditions the decomposition o

Alex787 [66]

Answer:

The rate equation for this reaction:

$R=k[NH_3]^0$

Explanation:

Decomposition of ammonia:

$2NH_3\rightarrow N_2+3H_2$

Rate law of the can be written as;

$R=k[NH_3]^x$

1) Rate of the reaction , when $[NH_3]=2.0\times 10^{-3} M$

$1.5\times 10^{-6} M/s=k[2.0\times 10^{-3} M]^x$ ..[1]

2) Rate of the reaction , when $[NH_3]=4.0\times 10^{-3} M$

$1.5\times 10^{-6} M/s=k[4.0\times 10^{-3} M]^x$ ..[2]

[1] ÷ [2]

$\frac{1.5\times 10^{-6}M/s}{1.5\times 10^{-6}M/s}=\frac{k[2.0\times 10^{-3}M]^x}{k[4.0\times 10^{-3}M]^x}$

On solving for x , we get ;

x = 0

The rate equation for this reaction:

$R=k[NH_3]^0$

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Answer:

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