HELP!! URGENT!! WILL GIVE BRAINIEST

What makes a solution a homogeneous mixture? Question 13 options:

Lesechka [4]

The answer is B. uniform in composition.

5 0

3 years ago

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the

Kamila [148]

The answer for the following problem is mentioned below.

Therefore the final temperature of the gas is 740 K

Explanation:

Given:

Initial pressure of the gas ( $P_{1}$ ) = 1.8 atm

Final pressure of the gas ( $P_{2}$ ) = 4 atm

Initial temperature of the gas ( $T_{1}$ ) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ( $T_{2}$ )

We know;

From the ideal gas equation;

we know;

P × V = n × R × T

So;

we can tell from the above equation;

P ∝ T

(i.e.)

$\frac{P}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

$P_{1}$ = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

$\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

$T_{2}$ = $\frac{333*4}{1.8}$

$T_{2}$ = 740 K

Therefore the final temperature of the gas is 740 K

8 0

3 years ago

Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y

denis23 [38]

Answer:

Volume of $CHCl_3$ = 15.31 mL

Volume of $CHBr_3$ = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of $CHCl_3$ = 1.492 g/mL

The density of $CHBr_3$ = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of $CHCl_3$ = P ml

and the volume of $CHBr_3$ = Q ml

the sum of their volumes should be equal to the total volume of the mixture

$P \ ml + Q \ ml = 20 ml$ ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

$\mathtt{(Density \ of \ CHCl_3 \times Vol. \ of \ CHCl_3 ) + (Density \ of \ CHBr_3 \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density \ of \ mixture \times volume \ of \ the \ mixture)}$

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g = 36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of $CHCl_3$ = 15.31 mL

Volume of $CHBr_3$ = 4.69 mL

6 0

3 years ago

What is the gram formula mass of Glycine , NH2CH2COOH

dem82 [27]

N=14
H= 1(5)= 5
C=12(2)=24
O=16(2)=32

= 75

6 0

2 years ago

Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o

vivado [14]

Answer:

For 1: The amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2: The amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

Explanation:

For 1:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of $KIO_3$ solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

$0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol$

Hence, the amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2:

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = $\frac{1}{2}\times 0.00025=0.000125mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

6 0

3 years ago