Into which two smaller groups are plants divided.

Estimate the ph of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (na2hpo4) and 1.25 mole of hyd

GenaCL600 [577]

The HCl added = 1.25 moles

and the moles of Na2HPO4 = 1 mole

Now when acid is added in the given solution of Na2HPO4

One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4

Na2HPO4 + H+ ---> NaH2PO4

Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution

So this will result into formation of a buffer of phosphoric acid and NaH2PO4

NaH2PO4 + H+ ---> H3PO4

pKa of H3PO4 = 2.1

so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58

so the pH will be in between 2.1 to 7.2

7 0

2 years ago

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→C

GREYUIT [131]

Answer:

$m=8.79kg$

Explanation:

First of all we need to calculate the heat that the water in the cooler is able to release:

$Q=\rho * V*Cp*\Delta T$

Where:

Cp is the mass heat capacity of water
V is the volume
$\rho$ is the density

$Q=1 g/cm^3 *15000 cm^3*4.184 \frac{J}{g*^{\circ}C}*(10-90)^{\circ}C$

$Q=-5020800 J=-5020.8 kJ$

To calculate the mass of CO2 that sublimes:

$-Q=\Delta H_{sub}*m$

Knowing that the enthalpy of sublimation for the CO2 is: $\Delta H_{sub}=571 kJ/kg$

$5020.8 kJ=571 kJ/kg*m$

$m=\frac{5020.8 kJ}{571 kJ/kg}=8.79kg$

6 0

2 years ago

What is the pH of a 3.40 mM of acetic acid, 500 mL in which 50 mL of 1.00 M NaOAc has been added? Ka HOAc is 1.8 x 10-5? What pe

Deffense [45]

Answer:

a) pH = 4.213

b) % dis = 2 %

Explanation:

Ch3COONa → CH3COO- + Na+

CH3COOH ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]

mass balance:

⇒ C CH3COOH + C CH3COONa = [ CH3COOH ] + [ CH3COO- ]

∴ C CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)

∴ C CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL

⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]

charge balance:

⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water

⇒ [ CH3COO- ] = [ H3O+ ] + 1.00

⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5

⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0

⇒ [ H3O+ ] = 6.12 E-5 M

⇒ pH = - Log [ H3O+ ] = 4.213

b) (% dis)* mol acid = C CH3COOH = 3.4

∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol

⇒ % dis = 3.4 / 1.7 = 2 %

7 0

2 years ago

What is the scaling factor if the molar mass of

Step2247 [10]

Answer:

The scaling factor is 5.

Explanation:

Hello there!

In this case, since the scaling factor is defined as the ratio of the molar mass of the molecular formula (complete) to the empirical formula (simplified), it is possible to compute it for the empirical formula of CH2O whose molar mass is 30 g/mol (12+2+16) as shown below:

$sf=150/30=5$

Therefore, we can also infer that the molecular formula would be:

$C_5H_1_0O_5$

Best regards!

6 0

2 years ago

2

Alina [70]

Answer:

N - 1s²2s²2p³

Explanation:

Nitrogen is located in the p-block of the periodic table (groups 13-18) and is on the 2nd period.

The 2nd period tells us the principal energy level (a quantum number) is n = 2. Therefore, it must have already filled up the 1s sublevel.

The groups 13-18 on period 2 tells us that the 2s sublevel is also filled.

Nitrogen is located in Group 15. That means that there are 3 electrons that have filled the 2p sublevel, out of a possible 6.

Therefore, our electron configuration is 1s²2s²2p³

2p³ (Shorthand Config)

[He] 2s²2p³ (Noble Gas Config)

7 0

3 years ago