What kind of object has an indefinite shape and indefinite volume? IH

The term _____ describes a substance that can act as both an acid and a base.

telo118 [61]

The term amphoteric describes a substance that can act as both an acid and a base.

8 0

3 years ago

If a chemical reaction such as photosynthesis begins with 6 atoms of carbon C , how many atoms of carbon C should be in the prod

lara [203]

Answer:

Option B. 6 atoms of carbon

Explanation:

In any chemical reaction, atoms from reactant side and product side must be the same.

This is photosynthesis reaction:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

8 0

3 years ago

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equiva

maxonik [38]

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of $HNO_3$

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of $C_6H_5NH_2$ .

$M_2\text{ and }V_2$ are the final molarity and volume of $HNO_3$ .

We are given:

$M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?$

Putting values in above equation, we get:

$0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL$

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of $C_6H_5NH_2$ + Volume of $HNO_3$

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

$\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=\frac{1}{2}[pK_w+pK_b+\log C]$

$pOH=\frac{1}{2}[14+4.87+\log (0.0437)]$

$pOH=8.76$

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24$

Thus, the pH of the solution is, 5.24

4 0

3 years ago

1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

Now put all the given values in above equation, we get:

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

Therefore, the number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

8 0

2 years ago

Convert the following with the correct number of significant figures:

larisa [96]

Answer:

1.78 × 10⁹ μg

Explanation:

We have to convert 1.78 kg to μg.

Step 1: Convert 1.78 kilograms to grams

We will use the conversion factor 1 kg = 10³ g.

1.78 kg × 10³ g/1 kg = 1.78 × 10³ g

Step 2: Convert 1.78 × 10³ grams to micrograms

We will use the conversion factor 1 g = 10⁶ μg.

1.78 × 10³ g × 10⁶ μg/1 g = 1.78 × 10⁹ μg

8 0

2 years ago