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Bogdan [553]
3 years ago
14

What is the conjugate acid base pair for H20+HBR->H30+BR

Chemistry
1 answer:
True [87]3 years ago
4 0

Answer:

TABLE OF CONJUGATE ACID-BASE PAIRS Acid BaseK a (25oC) HClO 4ClO 4 – H 2 SO 4HSO 4 – HCl Cl– HNO 3NO 3 – H 3 O +H 2 O H 2 CrO 4HCrO 4 –1.8 x 10–1

Explanation:

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When the Earth rotates, what happens to the constellations?
kobusy [5.1K]

Answer:

Constellations Changing Positions!!!!       :D  <----(smiley face)

Explanation:

Due to the earth's rotation, stars appear to move. As the Earth rotates from west to east, the stars appear to rise in the East, moving across south to set in the west. The Sun will appear to move through the stars, making one complete circuit of the sky in 365 days!!

(yes i'm literally 9+6 years old and idek why i'm doing this XD )

3 0
2 years ago
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Consider the following reversible reaction. 2H2O(g)&lt;—&gt;2H2(g)+O2(g) What is the equilibrium constant expression for the giv
masha68 [24]

The reaction: 2H2(g) + O2(g) → 2H2O(g), can be interpreted as: a. 2 moles of hydrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of water.

7 0
3 years ago
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Which of the following are examples of chemical changes? Select all that apply.
Natali5045456 [20]

Answer:

b. milk spoiling and c. firecrackers exploding

Explanation:

These are both chemical changes, the composition of them change when this happens and it cannot be reversed

6 0
2 years ago
Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th
aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = ?

Depression in freezing point = \Delta T_f

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample  = 0.275 mol/kg

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  non electrolyte)

\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg

\Delta T_f=11^oC

\Delta T_f=T- T_f

T_f=T- \Delta T_f=179^oC-11^oC=168^oC

168°C is the melting point of your impure sample.

4 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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