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3 years ago

Consider the total ionic equation below.

Chemistry

1 answer:

LenaWriter [7]3 years ago

7 0

Answer:

The spectator ions in this equation are :

CrO²⁻₄ and Ba²⁺

Explanation:

Spectator ions : These are the ions which are present in the solution but do not take part in chemical reaction.They are present as such in both reactant as well as product .

In this reaction,

1.H+ and OH- are combined and converted into water (H2O) : Changed to H2O

So , H+ and OH- are not Spectator ions

2.CrO²⁻₄ and Ba²⁺ are unchanged . Present as such in product and reactants

CrO²⁻₄ and Ba²⁺are Spectator ions

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A brick has a mass of 4.0 kg and the Earth has a mass of 6.0 × 1027 g. Use this information to answer the questions below. Be su

Vedmedyk [2.9K]

Answer:

a) $2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

b) $2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

Explanation:

a) Mass of brick = 4.0 kg

1 mole = $N_A=6.022\times 10^{23}$ particles/ atoms/molecules

Mass of $N_A$ bricks :

$=6.022\times 10^{23}\times 4.0 kg=2.4088\times 10^{24} kg\approx 2.4\times 10^{24} kg$

$2.4\times 10^{24} kg$ is the mass of 1 mole of bricks.

Mass of the Earth = M = $6.0\times 10^{27} kg$

Mass of 1 mole of brick = m= $2.4\times 10^{24} kg$

Let the moles of brick with equal mass of the Earth be x.

$m\times x=M$

$x=\frac{M}{m}=\frac{6.0\times 10^{27}kg}{2.4\times 10^{24} kg}=2.5\times 10^3$

$2.5\times 10^3$ moles of bricks have a mass equal to the mass of the Earth.

7 0

4 years ago

PLEASE PLEASE HELP!

valentinak56 [21]

Answer: The number of grams of $H_2$ in 1620 mL is 1.44 g

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$

$n=\frac{PV}{RT}$

$n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles$

Mass of hydrogen = $moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g$

The number of grams of $H_2$ in 1620 mL is 1.44 g

8 0

3 years ago

A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a

Mandarinka [93]

There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA]

where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87

6 0

3 years ago

Please help!!!!!!!

wolverine [178]

The answer is c. 104 g because 152-64 equals 88 and 192-88 equals 105

6 0

3 years ago

A salt contains only barium and one of the halide ions. A 0.158 g sample of the salt was dissolved in water, and an excess of su

Delvig [45]

Answer:

BaBr2

Explanation:

Halide ions are the ions of the elements of group 17 (or 7A), which have 7 electrons in their valence shell, so they need to gain one electron to be stable. If we noted as X the halide, the ion must be $X^{-}$ .

Barium is a metal of group 2 (or 2A) and has 2 electrons in its valence shell, so it must lose these 2 electrons to be stable, and it will form the anion $Ba^{2+}$ . These elements will form an ionic bond, with formula BaX2.

Sulfuric acid has formula H2SO4, and the reaction is:

BaX2 + H2SO4 --> BaSO4 + 2HX

So, for the stoichiometry between BaX2 and BaSO4, and denoting by <em>y</em> the molar mass of the halide, and the molar mass of BaSO4 = 137.33 g of Ba + 32.06 g of S + 4x16 g of O = 233.39

1 mol of BaX2 ------------- 1 mol of BaSO4

(137.33 + 2y) g of BaX2 ------------ 233.39 g of BaSO4

0.158 g of BaX2 --------------------- 0.124 g of BaSO4

By a simple direct three rule:

0.124x(137.33 + 2y) = 0.158x233.39

17.02892 + 0.248y = 36.87562

0.248y = 19.8467

y = 19.8467/0.248

y = 80.03 g/mol

By consulting the periodic table, the halide element that has a molar mass close to 80.03 g/mol is Bromide (Br), which has molar mass 79.9 g/mol. So the salt has formula BaBr2.

8 0

3 years ago