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4 years ago

Which of the following statements is not true regarding acids?

2 answers:

8 0

Active metals will react with acids in double replacement reactions.

NH4+ 

increase the volume 
add a catalyst

aev [14]4 years ago

6 0

The correct answer is the last option. Active metals will not react with acids in double replacement reactions. Instead, it reacts with metals via sing replacement reactions. It will displace hydrogen from the acid.

You might be interested in

What is the pH of 0.26 M ammonium ion? NH4+(aq) + H2O(1) NH3(aq) + H30* (aq) a. 4.33 b.9.25 c. 3.87 d. 4.92 e. 4.75

choli [55]

Answer:

b) pH = 9.25

Explanation:

NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
NH3 + H2O ↔ NH4+ + OH-
2 H2O ↔ H3O+ + OH-

⇒ Kb = [ NH4+ ] * [ OH- ] / [ NH3 ] = 1.86 E-5......from literature

mass balance NH4+:

⇒ M NH4+ = [ NH4+ ] - [ OH- ]

∴ [ NH3 ] ≅ M NH4+ = 0.26 M

⇒ Kb = (( 0.26 + [ OH- ] )) * [ OH- ] / 0.26 = 1.86 E-5

⇒ 0.26 [ OH-] + [ OH- ]² = 4.836 E-6

⇒ [ OH- ]² + 0.26 [ OH- ] - 4.836 E-6 = 0

⇒ [ OH- ] = 1.859 E-5 M

⇒ pOH = - Log ( 1.859 E-5 )

⇒ pOH = 4.7305

⇒ pH = 14 - pOH = 9.269

6 0

3 years ago

Il See Periodic Table See Hint

zavuch27 [327]

The molecular formula shows the number of atoms present. The molecular formula of the gas is most likely ClO2.

In terms of gas density and molar mass, the ideal gas equation can be written in the form; PM = dRT

Where;

P = pressure of the gas

M = molar mass of the gas

d = density of the gas

R = molar gas constant

T = temperature of the gas

Making the molar mass of the gas the subject of the formula;

M = dRT/P

d = 2.875 g/L

R = 0.082 atmLmol-1K-1

T = 11°C + 273 = 284 K

P = 750.0 mm Hg or 0.99 atm

Substituting values;

M = 2.875 g/L × 0.082 atmLmol-1K-1 × 284 K/ 0.99 atm

M = 67.6 g/mol

The gas is most likely ClO2.

Learn more: brainly.com/question/11969651

6 0

2 years ago

what is the formula equation for the reaction between sulfuric acid and dissolved sodium hydroxide if all products and reactants

Rama09 [41]

This reaction would produce salt and water- Sodium Sulphate and Water.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

4 0

3 years ago

Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to

BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = 0,8 M

[B] =0,2 M

[C] = 0,2M

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = $\frac{[C]^2}{[A][B]}$

a) The concentrations of each compound are:

[A] = $\frac{1,60 moles}{4,0 L}$ = 0,4 M

[B] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

[C] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

kc = $\frac{[0,1]^2}{[0,4][0,1]}$ = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = $\frac{1,60-x moles}{4,0 L}$

[B] = $\frac{0,60-x moles}{4,0 L}$

[C] = $\frac{0,60+2x moles}{4,0 L}$

0,25 = $\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}$

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119 → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

$\frac{1,60-(-0,04 moles)}{4,0 L}$ = 0,41 M

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = $\frac{1,60 moles}{2,0 L}$ = 0,8 M

[B] = $\frac{0,40 moles}{2,0 L}$ = 0,2 M

[C] = $\frac{0,40 moles}{2,0 L}$ = 0,2M

I hope it helps!

8 0

4 years ago

Read 2 more answers

GOOD REWARD I will report those who gives random answers.

Helen [10]

Answer:

3 mol AlCl₃.

Explanation:

Hello!

In this case, according to the specified reactants and products, it is possible to set up the following balanced chemical reaction:

$Al(NO_3)_3+3NaCl\rightarrow 3NaNO_3+AlCl_3$

Whereas we evidence the 1:3 mole ratio between aluminum nitrate and sodium chloride; thus, since different moles were reacting, we need to identify the limiting reactant by computing the moles of AlCl₃ produced by each reactant as follows:

$n_{AlCl_3}^{by\ Al(NO_3)_3}=4molAl(NO_3)_3*\frac{1molAlCl_3}{1molAl(NO_3)_3} =4molAlCl_3\\\\n_{AlCl_3}^{by\ NaCl}=9molNaCl*\frac{1molAlCl_3}{3molNaCl} =3molAlCl_3$

Thus, we infer that NaCl is the limiting reactant as it produces the fewest moles of AlCl₃; consequently the produced amount of this product is 3 mol.

Best regards!

5 0

3 years ago