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3 years ago

Ca(s)+br2(l)⟶cabr2(s) express your answer as a chemical equation. identify all of the phases in your answer.

1 answer:

3 0

The question requires to balance the equation.

The equation is already balanced, so the answer is:

Ca(s)+Br2(l)⟶CaBr2(s)


Explanation:


1) The phases are identified with a letter to the right of the compound or element:


So, for Ca, the phase is (s) which means solid.


For Br₂ (l) the phase is (l) which means liquid.


For CaBr₂(s) the phase is (s) which means solid.


2) The core work of balancing is make the number of atoms of each element on the reactants equal to the same number of atoms on the products side.


That is the law of conservation of mass applied to chemcial reactions.


3) So, you have to add coefficientes on the right place to make the number of atoms on the left side equal to the number of the same kind of atom on the right side.


In this table you can verify that the equation given is balanced:


atom left side right side

Ca 1 1

Br 2 2

So, you do not need to modify any coefficients.

You might be interested in

Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)

MrRissso [65]

Answer:

Equilibrium constant expression for $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq) \rightleftharpoons H_2CO_3\, (aq)$ :

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}$ .

Where

$a_{\mathrm{H_2CO_3}}$ , $a_{\mathrm{H^{+}}}$ , and $a_{\mathrm{CO_3}^{2-}}$ denote the activities of the three species, and
$[\mathrm{H_2CO_3}]$ , $\left[\mathrm{H^{+}}\right]$ , and $\left[\mathrm{CO_3}^{2-}\right]$ denote the concentrations of the three species.

Explanation:

<h3>Equilibrium Constant Expression</h3>

The equilibrium constant expression of a (reversible) reaction takes the form a fraction.

Multiply the activity of each product of this reaction to get the numerator. $\rm H_2CO_3\; (aq)$ is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be $\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)$ .

Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient " $2$ " on the product side of this reaction. $\rm 2\; H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ is equivalent to $\rm H^{+}\, (aq) + H^{+}\, (aq) + {CO_3}^{2-}\, (aq)$ . The species $\rm H^{+}\, (aq)$ appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:

$\left(a_{\mathrm{H^{+}}}\right)\cdot \left(a_{\mathrm{H^{+}}}\right)\cdot \, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right = \left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}})\right$ .

That's where the exponent " $2$ " in this equilibrium constant expression came from.

Combine these two parts to obtain the equilibrium constant expression:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}$ .

<h3 /><h3>Equilibrium Constant of Concentration</h3>

In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous " $(\rm aq)$ " species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:

$\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}$ .

8 0

3 years ago

I want to know which ones are molecular equation, complete ionic equation and net ionic equation

NNADVOKAT [17]

Answer:

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

8 0

3 years ago

Which gas is most abundant in Earth’s atmosphere?

natima [27]

Answer:

nitrogen

Explanation:

The atmosphere contains many gases, most in small amounts, including some pollutants and greenhouse gases. The most abundant gas in the atmosphere is nitrogen, with oxygen second. Argon, an inert gas, is the third most abundant gas in the atmosphere. Why do I care?

4 0

3 years ago

Read 2 more answers

Match the words in the below to the appropriate blanks in the sentences.

12345 [234]

Answer:

Sn and Ge

Explanation:

To obtain the more metallic element, we compare the group in which both elements are. Generally the element with the lower ionzation energy is he more metallic one.

Ionization energy increases fro left to right across a period. Ionization energy decreases down the group.

1. When comparing the two elements A s and S n , the more metallic element is ______based on periodic trends alone.

Sn has a lower ionization energy so it is the more metallic one.

2. When comparing the two elements G e and S b , the more metallic element is ________ based on periodic trends alone.

Ge has a lower ionizaiton energy compared to Sb. So it is more metallic element than Sb.

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4 years ago

Which of the following turns blue with a universal indicator (Litmus paper)

Aleks04 [339]

base

This is the answer

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3 years ago