Give an example of two elements that would form a covalent bond

The volume of a sample of oxygen is 300.0 mL, when the pressure is 1.00 atm and the temperature is 300K. At what pressure will t

Brut [27]

Answer:

The new pressure is 0.5 atm

Explanation:

Step 1: Data given

Volume of oxygen = 300 mL = 0.300 L

Pressure = 1.00 atm

Temperature = 300 K

The volume increases to 1000mL = 1.00 L

The temperature increases to 500 K

Step 2: Calculate the new pressure

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure = 1.00 atm

⇒with V1 = the initial volume = 0.300 L

⇒with T1 = the initial temperature = 300 K

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the increased volume = 1.00 L

⇒with T2 = the increased temperature = 500 K

(1.00 atm* 0.300 L)/300 K = (P2 * 1.00L) / 500 K

P2 = (1.00 *0.300 * 500) / (300 *1.00)

P2 = 0.5 atm

The new pressure is 0.5 atm

4 0

2 years ago

When the equation Al2(SO4)3 + NaOH + Al(OH)3 + Na2SO4 is correctly balanced, what is the coefficient of Al(OH)3?

nevsk [136]

The balanced chemical formula should be Al2(SO4)3 + 6NaOH = 2Al(OH)3 + 3Na2SO4

Therefore the coefficient of Al(OH)3 is 2!

Hope that helps :)

4 0

2 years ago

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!

3 0

3 years ago

Predict the number of peaks that you would expect in the proton-decoupled 13C spectrum of each comound.

dangina [55]

Answer:

To interpret a 13C-NMR spectrum we will use some standards very simple. A 13C-NMR spectrum gives us the following information:

1. Indicates the number of non-equivalent carbons in the molecule.

2. Measuring the chemical shift we can intuit the environment

electronic and determine the next functional groups.

3. In this case we cannot count on integration since the different

carbons have different relaxation times.

The number of peaks in the spectrum indicates the number of types of carbon present in the analyzed substance.

The factors that influence the chemical shift of the signals in the 13C NMR are:

electronegativity of carbon bound groups
carbon hybridization

Explanation:

The nuclear magnetic resonance of C13 is complementary to that of H1. This technique is used to determine the magnetic environment of carbon atoms.

4 0

3 years ago

Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4

Katena32 [7]

<u>Answer:</u> The mass of rust that can be removed is 1.597 grams

<u>Explanation:</u>

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = $6.00\times 10^2mL$ = 600 mL = 0.600 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol$

For the given chemical reaction:

$Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)$

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = $\frac{1}{6}\times 0.06=0.01mol$ of ferric oxide (rust)

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

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

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

$0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g$

Hence, the mass of rust that can be removed is 1.597 grams

5 0

3 years ago