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3 years ago

An example of a covalently bonded molecular formula for a diatomic molecule is

Chemistry

1 answer:

Westkost [7]3 years ago

5 0

Answer:

b. I2

Explanation:

I-I is covalently bonded and also diatomic , that's y!

✌️:)

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How many molecules are in a 1.5 mol sample of sodium oxide?

zzz [600]

Answer:

The numer of molecules is 9, 03 x 10 ^ 23 molecules. See the explanation below, please.

Explanation:

We use the number of Avogadro:

1 mol------- 6, 02x 10 10 ^ 23 molecules

1, 5mol ---x = (1, 5 mol x6, 02x 10 10 ^ 23 molecules)/ 1 mol=

x= 9, 03 x 10 ^ 23 molecules

7 0

3 years ago

BALANCING CHEMICAL EQUATIONS WORKSHEET

kherson [118]

Answer:

Put that in y butt ps

Explanation:

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3 years ago

A sample of nitric acid contains both H3O ions and NO3 ions. This sample has a pH

bearhunter [10]

Answer : The name of positive ion present in this sample is, hydronium ion.

Explanation :

For formation of a neutral ionic compound, the charges on cation and anion must be balanced.

The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

As per question, hydronium ion or hydrogen ion is having an oxidation state of +1 called as $H_3O^{+}$ or $H^+$ cation and nitrate ion $NO_3^{-}$ is an anion with oxidation state of -1.

Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $HNO_3$ .

Hence, the name of positive ion present in this sample is, hydronium ion.

4 0

4 years ago

When an electron loses electrons it becomes more 1 negative. 2 none of the above. 3 neutral 4 positive.

12345 [234]

The answer would be more positive

8 0

4 years ago

Read 2 more answers

Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M bu

Varvara68 [4.7K]

<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

<u>Explanation:</u>

The chemical equation for the dissociation of butanoic acid follows:

$CH_3CH_2CH_2COOH\rightleftharpoons CH_3CH_2CH_2COO^-+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}$

We are given:

$[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}$

$[CH_3CH_2CH_2COO^-]=[H^+]$

Putting values in above expression, we get:

$1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}$

$[H^+]=-0.0037,0.0037$

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is butanoic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=\frac{0.0037}{2}M\text{ (half equivalence)}\\V_1=20.00mL\\n_2=1\\M_2=0.425M\\V_2=?mL$

Putting values in above equation, we get:

$1\times \frac{0.0037}{2}\times 20.00=1\times 0.425\times V_2\\\\V_2=\frac{1\times 0.0037\times 20}{1\times 0.425\times 2}=0.087mL$

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

5 0

3 years ago