All categories

3 years ago

Could someone please help on just question 2?

Chemistry

1 answer:

sp2606 [1]3 years ago

4 0

Hey,

A compound is a substance that is made of two or more type of atoms chemically bonded together.
So this means that they are joined together by a chemical reaction.
And of those options PF3 H2SO4 H2O And CO are compounds as you can see they are made of two different type of atoms

Hope this helps !!

You might be interested in

A serving of Cheez-Its releases 1.30 x 10^4 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy we

kondor19780726 [428]

Answer:

The final temperature is:- 7428571463.57 °C

Explanation:

The expression for the calculation of heat is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$Q$ is the heat absorbed/released

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of water = 1.75 mg = 0.00175 g ( 1 g = 0.001 mg)

Specific heat of water = 4.18 J/g°C

Initial temperature = 35 °C

Final temperature = x °C

$\Delta T=(x-35)\ ^0C/tex]
Q = [tex]1.3\times 10^4$ kcal

Also, 1 kcal = 4.18 kJ = $4.18\times 10^3$ J

So, Q = $1.3\times 10^4\times 4.18\times 10^3$ J = 54340000 J

So,

$54340000=0.00175\times 4.18\times (x-35)$

$0.00175\times \:4.18\left(x-35\right)=54340000$

$x-35=\frac{54340000}{0.007315}$

$x=7428571463.57$

Thus, the final temperature is:- 7428571463.57 °C

3 0

2 years ago

Who conducted experiments to determine the quantity of charge carried by an electron?

MAXImum [283]

Answer:

it was Millikan. He conducted the oil drop experiment. Thomson determined the electon charge not the quantity. Rutherford used the gold foil experiment to find positive charge and that most of the atom is empty space. Dalton proposed that matter was made of small particles called atoms but that was a concept already proposed by ancient greeks. Dalton also proposed the atomic theory.

4 0

2 years ago

Balance the equation 2Fe + 3Cl2 → 2FeCl3

Tresset [83]

Answer:

6 Cl0 + 6 e- → 6 Cl-I (reduction)

2 Fe0 - 6 e- → 2 FeIII (oxidation)

4 0

2 years ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

3 years ago

Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks

charle [14.2K]

Answer:

pH = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15

Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15

K = Ksp X Kf

= 3*2*10^-15 * 10^15

= 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)

Initial: 0 0.3 0

Change: -2x +x

Equilibrium: 0.3 - 2x x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 -(2*0.089)

= 0.122

pOH = -log[OH⁻]

= -log 0.122

= 0.91

pH = 14-0.91

= 13.09

4 0

3 years ago