The orbits of ___ lie closer to the Sun than does Earth's orbit.

How many moles are in 3.90 x1030 atoms of Zn?

zaharov [31]

Explanation:
The problem basically wants you to find a way to convert between the number of atoms present in the sample and the number of moles they are equivalent to.
To convert between atoms and moles we use something called Avogadro's constant, which basically acts as the definition of a mole.
More specifically, in order to have one mole of an element you need
6.022 x 10^23 atoms of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.
In your case, you will have

3.90 x 10^ 26 atoms Zn x 1 mole Zn ( Avogrado’s constant) / 6.022 x 10^23 atoms Zn

= 6.5 x 10^8 is the answer

7 0

2 years ago

How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mmhg and a temperature of 21.0°c, can be made

Shalnov [3]

The answer is V = 0.6 L

The explanation:

1- First we will get P (H) :

p (H) = p total - p (H2O)

= 752 mmHg - 18.65 mmHg

= 733 mmHg

= 733 mmHg X (1 atm / 760 atm) = 0.964 atm

2- then when the reaction equation is:

Zn(s)+2HCl(aq) -> H2(g)+ZnCl2(aq)

so, then 1 mole of Zn will produce 1 mole of H2.

so, we need to get first the mole of Zn:

mole Zn = mass / molar mass

= 1.566g / 65.38g/mol

= 0.02395mol

∴ 0.02395 mol of Zn will produce 0.02395 mol of H2

by using the ideal gas equation we can get the volume V :

when :

PV = nRT

when T is the temperature = 21 + 273 = 294 K

and R is the ideal gas constant = 0.0821 L atm /K mol

and n is the number of moles = 0.2395 mol

and P is the pressure = 0.964 atm

so, by substitution:

V = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm)

= 0.6 L

3 0

2 years ago

Read 2 more answers

Which of these is an element? A. carbon B. carbon dioxide C. air D. water

Kaylis [27]

Elements are pure substances. Carbon is the only one that is pure.

4 0

3 years ago

Read 2 more answers

A. 14.8

Scilla [17]

Answer:

Explanation:

1)

Given data:

Initial volume = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

2)

Given data:

Initial volume = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1 mL

3)

Given data:

Initial volume = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75 L

T₂ = 720.78 L.K / 2.75 L

T₂ = 262.1 K

4)

Given data:

Initial volume = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89 L

5)

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8 L

6)

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79 mL

6 0

3 years ago

Which of the following substance is being reduced in the following reaction.? Cu (s) + 2AgNO3 (aq) + 2Ag (s) + Cu(NO3)2 (aq) A)

nikklg [1K]

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)$

The half reactions for the above reaction are:

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$

Reduction half reaction: $2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)$

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized.

Silver ion is gaining electrons and thus is getting reduced.

Hence, the correct answer is Option A.

3 0

2 years ago