Answer:
72.0 mL of steam is formed.
Explanation:
The reaction is :

You can treat coefficient of compounds as amount of volume used.
Therefore for 4 mL of ammonia 5 mL of oxygen is used to form 4 mL of nitric oxide gas and 6 mL of steam.
For 1 mL of ammonia
(=1.25) mL of oxygen is used to form
(=1) mL of nitric oxide gas and
(=1.5) mL of steam.
OR
Just transform the chemical equation by dividing the whole equation by 4 so that the coefficient of
become one like this

We don't know which one will be completely exhausted and which one will be left so we have to consider two cases :
<em>1. </em><em>Assume ammonia to be completely exhausted</em>
For 50 mL of ammonia
(= 62.5) mL of oxygen is needed. But we have just 60 mL of oxygen so this assumption is false.
2. <em>Assume oxygen to be completely exhausted</em>
For 60 mL of oxygen only
(=48) mL of ammonia is needed. In this case we have sufficient amount of ammonia. So this case is true.

Now we know that during complete reaction 48 mL of ammonia and 60 mL of oxygen is used which will form
(= 48) mL of nitic oxide gas and
(= 72) mL of steam.
Therefore <em>72 mL of steam </em>is formed.
Answer:
a) 10.0 mm
b) 8.7 x 10¹³ times
Explanation:
Atom diameter = 1.06 x 10⁻¹⁰ m ________________ 100%
Nucleus diameter = 2.40 x 10⁻¹⁵ m ______________ x
x = 2.26 x 10⁻³ %
The nucleus diameter is equivalent to 2.26 x 10⁻³ % of the total atom size.
a) The Empire State Building model:
1 ft = 304.8 mm
1454 ft = 443179.2 mm
443179.2 mm______ 100%
y ______ 2.26 x 10⁻³ %
y = 10.0 mm
In this model, the diameter of the nucleus would be 10.0 mm.
b) Sphere volume: V =(4 · π · r³
)/3
V atom = (4 . π .( 0.53x10⁻¹⁰)³ )/3
V atom = 6.2 x 10⁻³¹ m³
V nucleus = (4 . π .( 1.2x10⁻¹⁵)³ )/3
V nucleus = 7.2 x 10⁻⁴⁵ m³
V atom / V nucleus = 6.2 x 10⁻³¹ m³ / 7.2 x 10⁻⁴⁵ m³
V atom / V nucleus = 8.7 x 10¹³
The atom is times 8.7 x 10¹³ larger in volume than its nucleus.
oxygen and silicon I believe
I'm pretty sure it's B because carbon atoms are in all living organisms. They can also be bonded in different varations.