Groups = columns while Periods = Rows true or false

Sergio039 [100]

Answer: is neither an acid nor a base

Explanation: Water is a universal solvent which means it can dissolve most of the substances in it.

Water has high thermal heat capacity , which means large heat is required to heat the water.

Water is not always pure as it gets contaminated by various pollutants present in the atmosphere such as gases, bacteria and suspended matter.

Water is an amphoteric substance which can act as both acid and base, thus can donate and acept [texH^+[/tex] ions.Thus it is neither an acid nor a base.

$HCl+H_2O\rightarrow H_3O^++Cl^-$

Here water is accepting a proton, thus it acts as base.

$NH_3+H_2O\rightarrow NH_4^++OH^-$

Here water is donating a proton, thus it acts as acid.

4 0

3 years ago

Joel has noticed that his engine is not compressing properly during the compression and ignition step. What problem will this po

denis-greek [22]

The answer is: The engine will run inefficiently APEX....

7 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

What is the electron structure of sodium

yarga [219]

Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹

8 0

4 years ago

Read 2 more answers

Which of the following is a challenge of prosthetic engineering that has been met through tissue engineering?

solniwko [45]

Answer:

biocompatibility

Explanation:

5 0

2 years ago