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kirill [66]
2 years ago
7

How many atoms of hydrogen are there in 18g of water ?

Chemistry
1 answer:
Bingel [31]2 years ago
6 0

Answer:

They are 1.204×10^24 atoms of hydrogen present in 18 grams of water. In order to calculate this,it is necessary to compute the number of hydrogen moles present in the sample.

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a 1.00litre vessel contains 0.215 mole of nitrogen gas and 0.0118 mole of hydrogen gas at 25°C. determine the partial pressure o
stepladder [879]

Answer:

THANK YOU SA BRAINLIEST

8 0
2 years ago
What is the difference between accuracy and precision in chemistry ?
earnstyle [38]
Accuracy is more of a range 
6 0
4 years ago
Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C
Zanzabum

Answer:

HSO_3^-: conjugate acid of SO_3^{2-}

CH_3NH_2 : conjugate base of CH_3NH_3^+

SO_3^{2-} : conjugate base of HSO_3^-

CH_3NH_3^+ : conjugate acid of CH_3NH_2

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.

HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+

Here in forward reaction CH_3NH_2 is accepting a proton, thus it is considered as a base and after accepting a proton, it forms CH_3NH_3^+ which is a conjugate acid.

And HSO_3^-  is losing a proton, thus it is considered as an acid and after loosing a proton, it forms SO_3^{2-} which is a conjugate base.

Similarly in the backward reaction, CH_3NH_3^+ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms CH_3NH_2 which is a conjugate base.

And SO_3^{2-}  is accepting a proton, thus it is considered as a base and after accepting a proton, it forms HSO_3^{-} which is a conjugate acid.

4 0
3 years ago
What happens in this reaction? <br><br> butan-1-amine + CH3I --&gt;
Galina-37 [17]

Here we have to get the product between the reaction of butane-1-amine with methyl iodide (CH₃I).

The  reaction between 1 mole of butan-1-amine and 1 mole of methyl iodide produces Methyl-butamine which is a secondary amine.

However, In presence of 2 moles of methyl iodide the reaction proceed to N, N-di-methylbutamine. The reaction is shown in the figure.

This is one of the effective reaction method to generate secondary and tertiary amine from primary amine.

The primary amine reacts with alkyl iodide to form secondary to tertiary amine. The final product depends upon the quantity of the alkyl iodide present in the reaction.

7 0
3 years ago
When the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor. (How would this be writt
strojnjashka [21]
<span>When the gases dihydrogen sulfide and oxygen react, they form the gases sulfur dioxide and water vapor. so the balanced equation is
</span>2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(l)
hope it helps
7 0
3 years ago
Read 2 more answers
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