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astra-53 [7]
2 years ago
14

The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil

limeters? A) 1.15x 10-8 mm B) 1.15 1013 mm C) 1.15 10-7 mm D) 1.15 x 1017 mm
Chemistry
1 answer:
jasenka [17]2 years ago
5 0

Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Given data

Average distance between nitrogen and oxygen atoms: 115 pm

Step 2: Convert the distance to meters (SI base unit)

We will use the conversion factor 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Convert the distance to millimeters

We will use the conversion factor 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

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A student wants to form 2‑hexanol using acidic hydration. He finds four alkenes in the inventory cabinet that could be possible
julia-pushkina [17]

Answer:

Look on the picture.

Explanation:

He could find only 2 isomers of n-hexane alkenes for this reaction. Other two could be marked from other direction.

8 0
2 years ago
Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3
ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×\frac{1molZn}{65,38g}×\frac{1molH_{2}(g)}{1molZn} = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×\frac{101,325}{1atmL} = <em>19,26 kJ</em>

I hope it helps!

5 0
2 years ago
The U.S. Mint produces a dollar coin called the American Silver Eagle that is made of nearly pure silver. This coin has a diamet
Rudik [331]

Answer:

The value of the silver in the coin is 35.3 $

Explanation:

First of all, let's calculate the volume of the coin.

2π . r² . thickness = volume

r = diameter/2

r = 41 mm/2 = 20.5 mm

2 . π . (20.5 mm)² .  2.5 mm = 6601 mm³

Now, this is the volume of the coin, so we must find out how many grams are on it.

6601 mm³ / 1000 = 6.60 cm³

Let's apply density.

D = Mass / volume

10.5 g/cm³ = mass /6.60 cm³

10.5 g/cm³ . 6.60 cm³ = mass

69.3 g = mass

Each gram has a cost of 0.51$

69.3 g . 0.51$ = 35.3 $

7 0
2 years ago
A chemist uses 0.25 L of 2.00 M H2SO4 to completely neutralize a 2.00 L of solution of NaOH. The balanced chemical equation of t
dalvyx [7]
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?

n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)

c₂v₂=2c₁v₁

c₂=2c₁v₁/v₂

c₂=2*2.00*0.25/2.00=0.5 mol/L

0.5 M NaOH


4 0
2 years ago
Read 2 more answers
Imagine you needed to identify if an object has undergone a physical change or a chemical change. What information would you nee
pishuonlain [190]
How it looks. basically the thing that tells you how it change. for example if an ice cube was melted (heat), it only changed physically not chemically as the h20 molecules are still there. however lets say you burn woos— you cant get that would back. its ash now and it has changed chemically.
7 0
2 years ago
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