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3 years ago

How many significant figures are in 0.0233?

Chemistry

1 answer:

viva [34]3 years ago

7 0

Answer:

there are 3 significant figures

Explanation:

do not count the first 2 zeros. only the nimbers after the zero

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Write the net ionic equation for the reaction between hydrocyanic acid and potassium hydroxide. Do not include states such as (a

eimsori [14]

Answer:

The net ionic equation is as follows:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

Explanation:

The reaction between Hydrocyanic acid, HCN, and sodium hydroxide is a neutralization reaction between a weak acid and a strong base.

Hydrocyanic acid being a weak acid ionizes only slightly, while sodium hydroxide being a strong base ionizes completely. The equation for the reaction is given below:

A. HCN(aq) + NaOH-(aq) ----> NaCN(aq) + H2O(l)

Since Hydrocyanic acid is written in the aqueous form as it ionizes only slightly and the ionic equation is given below:

HCN(aq) + Na+(aq)+OH-(aq) ----> Na+(aq)+CN-(aq) + H2O(l)

Na+ being a spectator ion is removed from the net ionic equation given below:

HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)

4 0

3 years ago

Approximately how many elements combine in different ways to make compounds?

Vlada [557]

Mos commonly 2 elements combine.

examples: H2, O2

8 0

2 years ago

Read 2 more answers

Select all that apply.

stich3 [128]

Answer:

air pollution and potential reactor accidents

Explanation:

i pretty sure on those two

4 0

2 years ago

Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

sergejj [24]

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$ .

Since, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of $Al(C_{2}H_{3}O_{2})_{3}$ .

Thus, according to molecular formula, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.

1 atom of oxygen will be present in $\frac{1}{3.61\times 10^{24}}$ moles of $Al(C_{2}H_{3}O_{2})_{3}$ . Thus,

$2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}$ .

Molar mass of $Al(C_{2}H_{3}O_{2})_{3}$ is 236 g/mol, mass can be calculated as follows:

$m=n\times M=0.7285 mol\times 236 g/mol=172 g$

Therefore, mass of $Al(C_{2}H_{3}O_{2})_{3}$ will be 172 g.

5 0

3 years ago

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A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the sa

nexus9112 [7]

For the answer to the question above, I can't help you directly because I don't have a calculator right now. But I'll show you how to solve this.
use the freezing point depression formula for this one: delta T = i * m * K where K is a constant, m is the molality (mol solute/kg solvent), and i is the van'hoff factor the van hoff factor is the number of ions that your salt dissociates into. Since it's an ALKALI flouride salt, how many ions? k is just a constant, you get it from a table in your textbook somewhere So you have everything to solve for the molality of the solution, once you did that, multiplying it by the mass of water to find the mols of the salt. Take the mass of the salt and divide by this mols to figure out the molar mass, and then compare it with the periodic table to identify the salt.

Mole solute x mass of Water = Mol solute
kg Solvent

then

Mass of solute x 1 = molar mass
mole of solute

4 0

2 years ago

Read 2 more answers