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3 years ago

Write the net ionic reaction for the following double replacement reaction:

Chemistry

1 answer:

Schach [20]3 years ago

6 0

Net ionic equation

Cu²⁺(aq)+S²⁻(aq)⇒CuS(s)

<h3>Further explanation</h3>

Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

In the ion equation, there is a spectator ion that is the ion which does not react because it is present before and after the reaction

When these ions are removed, the ionic equation is called the net ionic equation

For gases and solids including water (H₂O) can be written as an ionized molecule

Reaction

CuSO₄(aq)+Na₂S(aq)⇒CuS(s)+Na₂SO₄

ionic equation

Cu²⁺(aq)+SO₄²⁻(aq)+2Na⁺(aq)+S²⁻(aq)⇒CuS(s)+2Na⁺(aq+SO₄²⁻(aq)

spectator ions : 2Na⁺ and SO₄²⁻

Net ionic equation

Cu²⁺(aq)+S²⁻(aq)⇒CuS(s)

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Given the following reaction: 3D(g) + E(g) + 2F(g) → 5G(g) + 4H(g)

evablogger [386]

Answer:

Explanation:

Hello,

The law of mass action, allows us to know the required amounts, thus, for this chemical reaction it is:

$\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}$

Now, we answer:

(a)

$\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s$

(b)

$\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s$

$r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}$

Thus, any of the available expressions are suitable to quantify the rate of the reaction.

Best regards.

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4 years ago

What process will decrease the level of co2 in the atmosphere

ratelena [41]

Photosynthesis will decrease the level of co2 in the atmosphere

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3 years ago

Which of the following describes a difference between galaxies and solar systems? A. Galaxies contain only moons, while solar sy

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4 years ago

Read 2 more answers

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0

grandymaker [24]

The given question is incomplete. The complete question is:

Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced

Answer: 60.0 g of glucose

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $CO_2$

$\text{Number of moles}=\frac{88.0g}{44g/mol}=2.0moles$

b) moles of $H_2O$

$\text{Number of moles}=\frac{64.0g}{18g/mol}=3.5moles$

$6CO_2+6H_2O\rightarrow C_{6}H_{12}O_6+6O_2$

According to stoichiometry :

6 moles of $CO_2$ require = 6 moles of $H_2O$

Thus 2.0 moles of $CO_2$ require= $\frac{6}{6}\times 2.0=2.0moles$ of $H_2O$

Thus $CO_2$ is the limiting reagent as it limits the formation of product.

As 6 moles of $CO_2$ give = 1 moles of glucose

Thus 2.0 moles of $CO_2$ give = $\frac{1}{6}\times 2.0=0.33moles$ of glucose

Mass of glucose = $moles\times {\text {Molar mass}}=0.33moles\times 180g/mol=60g$

Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water

8 0

3 years ago

Write the beta decay equation for the following isotope: 91 38 Sr? Please write out steps

babymother [125]

Answer:

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e + \, _{39}^{91}Y$

Explanation:

The unbalanced nuclear equation is

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e \, + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{38}^{91}Se} \longrightarrow \, _{-1}^{0}e \, + \, _{Z}^{A}X$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

98 = 0 + A, so A = 98 - 0 = 98, and

38 = -1 + Z, so Z = 38 + 1 = 39

Then, your nuclear equation becomes

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e + \, _{39}^{91}X$

Element 39 is yttrium, so the balanced nuclear equation is

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e + \, _{39}^{91}Y$

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3 years ago