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2 years ago

Michelle has a beaker containing NaCl. What is in Michelle's beaker

Chemistry

2 answers:

dem82 [27]2 years ago

4 0

Her beaker contains sodium chloride.

True [87]2 years ago

3 0

<span>Sodium chloride is in her beaker </span>

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Y_Kistochka [10]

Answer:

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3 years ago

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Iodine-138 decays by beta decay. What element does it produce?

iVinArrow [24]

${xe}^{138}$

Explanation:

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2 years ago

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Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ

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2 years ago

H2CO3(aq) + H200 H30 (aq) + HCO3 (aq).

timofeeve [1]

Answer:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

an equilibrium constant is, first of all, a fraction;
in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
each concentration should be raised to the power of the coefficient in the balanced chemical equation;
only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

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