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2 years ago

Sodium carbonate + hydrochloric acidsodium chloride + carbon dioxide + water

Chemistry

1 answer:

Minchanka [31]2 years ago

7 0

Answer:

a carbonate reacts with an acid to form a salt,water and carbon(iv) oxide

Explanation:

note the products well

ex:acid and base=salt and water only

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Calculate the percentage of water of crystallisation in MgSO⁴ 7H²O

Bas_tet [7]

Answer:

Formula of EPSOM salt = MgSO4.7H2O

molecular mass of MgSO4.7H2O = atomic mass of Mg + atomic mass of S + 4 × atomic mass of O + 7 { 2 × atomic mass of H + atomic mass of O }

= 24 + 32 + 4× 16 + 7{ 2 × 1 + 16 } g/mol

= (24 + 32 + 64+ 126 ) g/mol

= 246 g/mol

molecular mass of total water = 7 × ( 2× atomic mass of H + atomic mass of O )

= 7 × 18 = 126 g/mol

now ,

% mass of H2O in EPSOM salt = {total molar mass of H2O/molar mass of Epsom salts }× 100

= {126/246 } × 100

= 12600/246

= 51.21 %

Explanation:

i have done it hope it helps

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2 years ago

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Hello please please help me

olga nikolaevna [1]

Answer:

Option-A is the correct answer

Explanation:

Lithium belong to group 1 metals. Hence, it can loose one electron to form lithium ion i.e. Li⁺¹ or Li⁺

While, Nitrogen is non-metal and hence has the ability to gain the electron lost by lithium metal. Furthermore, Nitrogen can gain maximum 3 electrons to acheive noble gas configuration. Hence, three Li atoms will loose their electrons and Nitrogen will gain those three electrons to form nitride ion i.e. N³⁻.

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2 years ago

What do scientist use to measure the volume of a substance

natali 33 [55]

Scientists use a Graduated Cylinder

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2 years ago

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Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. Include all phases in answer

liq [111]

MnO2(s)+H2(g)→MnO(s)+H2O(g)

4 0

2 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago