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UNO [17]
3 years ago
14

BaCl2(aq) + Na2CO3(aq) BaCO3(s) + NaCl(aq

Chemistry
1 answer:
Vitek1552 [10]3 years ago
7 0
This is the equation balanced:

<span>BaCl2(aq) + Na2CO3(aq) = BaCO3(s) + 2 NaCl(aq)

Then the coefficient in front of Na Cl is 2.

Answer: 2
</span>


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MARKING BRAINLIEST! please help asap, i need a and b, thank you
defon

The time required to reduce the concentration from 0.00757 M to 0.00180 M is equal to 1.52 × 10⁻⁴ s. The half-life period of the reaction is 9.98× 10⁻⁵s.

<h3>What is the rate of reaction?</h3>

The rate of reaction is described as the speed at which reactants are converted into products. A catalyst increases the rate of the reaction without going under any change in the chemical reaction.

Given the initial concentration of the reactant, C₀= 0.00757 M

The concentration of reactant after time t is C₁= 0.00180 M

The rate constant of the reaction, k = 37.9 M⁻¹s⁻¹

For the first-order reaction: C_t =C_0-kt

0.00180 = 0.00757 - (37.9) t

t =  1.52 × 10⁻⁴ s

The half-life period of the reaction: t_{\frac{1}{2} } =\frac{C_0}{2k}

t_{\frac{1}{2} } =\frac{0.00757}{2\times 37.9}

Half-life of the reaction = 9.98 × 10⁻⁵s

Learn more about the rate of reaction, here:

brainly.com/question/13571877

#SPJ1

5 0
10 months ago
A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a
Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄  

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

                 BaC₂O₄ ⇌   Ba²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                   5.0 × 10⁻⁵     c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

                ZnC₂O₄ ⇌   Zn²⁺   + C₂O₄²⁻

E/mol·L⁻¹:                 2.0 × 10⁻⁷      c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

                 Ag₂C₂O₄ ⇌   2Ag⁺   +   C₂O₄²⁻  

E/mol·L⁻¹:                      3.0 × 10⁻⁵       c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when   c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when   c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028       mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0
3 years ago
(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

5 0
3 years ago
What is the mass of HCL is consumed by the reaction of 1.5 mole of magnesium
photoshop1234 [79]

Answer: 109.5g

Explanation:

Mg + 2HCl —> MgCl2 + H2

From the equation,

1mole of Mg required 2moles of HCl

Therefore 1.5moles of Mg will require = 1.5 x 2 = 3 moles of HCl.

Molar Mass of HCl = 1+35.5 =36.5g/mol

Mass conc. Of HCl = 3 x 36.5 = 109.5g

3 0
2 years ago
The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 9.00 g
timofeeve [1]

Answer:

20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

Explanation:

Heat is being consumed during vaporization and heat is being released during condensation.

To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.

Molar mass of water = 18.02 g/mol

Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.

So, to vaporize 9.00 g of water, (\frac{40.66}{18.02}\times 9.00)kJ of heat or 20.3 kJ of heat is being consumed

As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

5 0
3 years ago
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