_____ energy is required to start a reaction

A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0

3 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

Predict whether ΔS° is greater than, less than, or approximately zero for each of the following reactions, and explain your choi

inna [77]

Answer:

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .

If in a reaction , more number of gaseous atoms are present in the product side , entropy will increase , i.e. Δ°S > 0

When liquid is converted to solid entropy decreases,

As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

If in a reaction , less number of gaseous atoms are present in the product side , entropy will decrease , i.e. Δ°S < 0

From the question ,

( a ) NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

Gaseous atoms -

Reactant - 1 + 5 = 6

Product - 4 + 6 = 10 ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

( b ) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

Gaseous atoms -

Reactant - 1 + 2 = 3

Product - 1 + 2 = 3 ,

Since ,

Both the side the value of gaseous atoms are , hence , Δ°S = 0 .

( c ) CaCO₃(s) → CaO(s) + CO₂(g)

Gaseous atoms -

Reactant = 0

Product - 0 + 1 = 1 ,

Since ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

8 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

dezoksy [38]

What is the question?

7 0

2 years ago

What is 5 ounces in milliliters?

SIZIF [17.4K]

1 ounces ----------- 29.5735 mL
5 ounces ----------- ??

5 x 29.5735 / 1 => 147.868 mL

hope this helps!

6 0

3 years ago