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lisov135 [29]
2 years ago
6

#4 only plss and huryy ​

Chemistry
1 answer:
scoray [572]2 years ago
6 0

Answer:C

Explanation: The mass is still the same even though they took it apart.

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Approximately 30% of the earth’s fresh water is stored in– A lakes. B streams. C groundwater. D living things.
kirill115 [55]
C ground water hope this is helpful
7 0
3 years ago
If you had an aqueous mixture that contained Ag+ , K+ , and Pb2+ cations, how many different solids could precipitate if a chlor
SCORPION-xisa [38]

Answer:

Two, KCl and PbCl₂.

Explanation:

Hello!

In this case, since the addition of chloride ions promote the following three ionic reactions:

Ag^+(aq)+Cl^-(aq)\rightleftharpoons AgCl(s)\\\\K^+(aq)+Cl^-(aq)\rightleftharpoons KCl(aq)\\\\Pb^{2+}(aq)+Cl^-(aq)\rightleftharpoons PbCl_2(s)

We can infer that both silver chloride and lead (II) chloride are precipitated products as their Ksp are 6.56x10⁻⁴ and 1.59x10⁻⁵ respectively, which means they are merely soluble in water.

Best regards!

3 0
3 years ago
A Hydrocarbon C₄H₁₀ has a boiling point of 0, another hydrocarbon C₆H₁₄ has a boiling point of 69. What would you expect the boi
cupoosta [38]
If the trend is linear, that is boiling point increases by a constant amount for each additional saturated carbon, the boiling point of octane would be 69*2 because you are adding two saturated carbons. So the predicted boiling point is 138C, which is very close to the literature value
5 0
2 years ago
Read 2 more answers
A 1. 07 g sample of a noble gas occupies a volume of 363 ml at 35°c and 678 mmhg. Identify the noble gas in this sample. (r = 0.
Margaret [11]

The identity of the noble gas is the sample is Krypton

<h3>Ideal Gas law</h3>

From the question, we are to determine the identity of the noble gas in the sample

From the ideal gas equation, we have that

PV = nRT

∴ n = PV / RT

Where P is the pressure

V is the volume

n is the number of moles

R is the gas constant

and T is the temperature

From the given information,

P = 678 mmHg = 0.892105 atm

V = 363 mL = 0.363 L

R = 0.08206 L.atm/mol.K

T = 35 °C = 35 + 273.15 K = 308.15 K

Putting the parameters into the equation, we get

n = (0.892105 × 0.363)/ (0.08206 × 308.15)

n = 0.0128 moles

Now, we will determine the Atomic mass of the sample

Using the formula,

Atomic = Mass / Number of moles

Atomic mass of the substance = 1.07 / 0.0128

Atomic mass of the substance = 83.6 amu

The noble gas with the closest atomic mass to this value is Krypton.

Molar mass of Krypton = 83.798 amu

Hence, the identity of the noble gas is the sample is Krypton

Learn more on Ideal Gas law here: brainly.com/question/20212888

#SPJ12

4 0
2 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
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