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erica [24]
3 years ago
6

How to balance MgCo3+HCL=Mgcl+H2o+ Co2​

Chemistry
1 answer:
algol133 years ago
3 0

Answer:

MgCO3 + 2HCl → MgCl2 + CO2 + H2O

Explanation:

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What would happen to a sealed bag of chips left in the sun?
olchik [2.2K]

The answer is the gaz inside the bag would expand. (A)

7 0
3 years ago
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The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in
pantera1 [17]

Answer:

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

Explanation:

Half life for second order kinetics is given by:

t_{\frac{1}{2}=\frac{1}{k\times a_0}

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

t_{\frac{1}{2} = half life

k = rate constant

a_0 = initial concentration

a = Final concentration of reactant after time t

We have :

C_2F_4 \longrightarrow \frac{1}{2} C_4F_8

Initial concentration of C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L

Rate constant = k = 0.041 M^{-1} s^{-1}

t_{\frac{1}{2}=\frac{1}{k\times a_0}

=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}

t_{1/2}=134.8 s

134.8 seconds is the half-life (in seconds) of the reaction for the initial C_2F_4 concentration

3 0
3 years ago
Suppose you have a mixture of solid mg(oh)2 and baso4, how do you separate the mixture
Step2247 [10]
This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.

Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O

The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.

MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄

Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
5 0
3 years ago
A gas with a pressure of 820.4 mmHg occupies a
irga5000 [103]

Answer:

V₂ = 1223.2 mL  

Explanation:

Given data:

Pressure of gas = 820.4 mmHg

Initial volume of gas = 900.0 mL

Initial temperature = 25.0°C (25+273=298K)

Final temperature = 132.0°C (132.0 +273 = 405 K)

Final volume = ?

Solution:

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 900.0 mL × 405 K / 298 k

V₂ = 364500 mL.K / 298 K

V₂ = 1223.2 mL  

5 0
3 years ago
A 1.00-molecule sample of glucose C6H12O6 completely reacts with oxygen as represented by the balanced equation below
olga_2 [115]
<span>C</span>₆<span>H</span>₁₂<span>O</span>₆ + 6 <span>O</span>₂ = 6 <span>CO</span>₂ + 6 <span>H</span>₂<span>O

</span>glucose + oxygen = Carbon Dioxide + Water

combustion <span>reaction
</span>
hope this helps!
3 0
3 years ago
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