Ask question

Ask question

All categories

anzhelika [568]

3 years ago

10

What is the Formula for sodium chlorate

2 answers:

FrozenT [24]3 years ago

7 0

T<span>he Formula for sodium chlorate is- </span>NaClO3.

Alex Ar [27]3 years ago

7 0

The formula for sodium chlorate is NaClO3

Hope I helped :)

You might be interested in

If a sodium atom looses an electron to form an ion what is its charge?

Igoryamba

Is there any answers to choose from?
<span />

7 0

3 years ago

Read 2 more answers

What's the IUPAC name pf this compound? CH3-CH-CH2-CH-CHO | | CH3 CL

Svetradugi [14.3K]

Answer:

2-chloro-4-methylpentanal.

Explanation:

Hello there!

In this case, according to the chemical compound:

CH3-CH-CH2-CH-CHO

| |

CH3 Cl

We can see the main functional group is an starting carbonyl, which means this is an aldehyde. Moreover, we can see a Cl-substituent on the second carbon and a methyl substituent on the fourth carbon. Therefore, the IUPAC name turns out: 2-chloro-4-methylpentanal.

Best regards!

8 0

3 years ago

Convert 5.50 mol of MgO to g.

Fiesta28 [93]

Answer:

5.50 moles of magnesium oxide is 221.6742 grams

Explanation:

to do this you multiply the number of moles by the molar mass

7 0

2 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

3 0

3 years ago

Identify whether the following example is qualitative or quantitative data.

OLga [1]

I think that it is qualitative data

8 0

3 years ago