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Kamila [148]
3 years ago
6

The function of hydrochloric acid in the gastric juice.

Chemistry
1 answer:
irina [24]3 years ago
5 0
The function of hydrochloric acid is to melt the food
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When an electron in an atom moves from a higher, less stable level, down to a lower, more stable level, a photon is
Wittaler [7]

Answer:

Emitted

Explanation:

8 0
3 years ago
For Mn3+, write an equation that shows how the cation acts as an acid.
Mkey [24]

An acid is a compound which will give H+ ions or H3O^+  ions

the reaction will be

[Mn(H_{2}O )_{6} ^{+3} +H_{2}O --> [MnOH(H_{2}O)_{5}]^{+2} + H_{3}O^{+}

Thus as there is evolution of H_{3}O^{+} the Mn+3 is an acid

8 0
3 years ago
Read 2 more answers
18.2L of gas at 95°C and 760 torr is placed in a 15L container at 80 degrees * C ; what is the new pressure ?
romanna [79]

Answer:

884.56 torr

Explanation:

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

P = Pressure

V = Volume

T = Temperature in kelvin (Celsius + 273.15)

\frac{(760)(18.2) }{368.15} = \frac{P(15) }{353.15}

P = \frac{(760)(18.2)(353.15) }{(368.15)(15)}

P = 884.56169

4 0
3 years ago
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0
3 years ago
6 moles of H2O is equal to how many molecules?
alexandr1967 [171]
6 miles of H2O is equal to 12 Hydrogen molecules and 6 oxygen molecules. Equaling 18 in total.
5 0
3 years ago
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