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A student has 5 mL of 0.300 M (NH4)2CrO4 solution. What is the concentration of NH4+ in this solution?

mixer [17]

The answer is 0.600 M; but I don't know how to get to that answer.

4 0

3 years ago

What is the first step of a ladybug during the growth and development process

Sunny_sXe [5.5K]

Answer:

Damian here! (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

The newly hatched larva is in its first instar, a developmental stage that occurs between molts. It feeds until it grows too big for its cuticle, or soft shell, and then it molts. After molting, the larva is in the second instar. Ladybug larvae usually molt through four instars, or larval stages, before preparing to pupate.

Explanation:

hope this helps? :))

7 0

3 years ago

Calcium cyclamate, Ca(C₆H₁₁NHSO₃)₂, is an artificial sweetener used in many countries around the world but is banned in the Unit

xeze [42]

Answer:

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃;

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃;

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

Explanation:

First, let's see the reactants for the first reaction and how they dissociate:

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

BaCO₃ → Ba²⁺ + CO₃²⁻ (Barium is from group 2, so its cation has charge +2)

So, to form the products, the cation of one will join the anion of others. The amount of the cation will be the charge of the anion, and the amount of the anion will be the charge of the cation:

H⁺ + CO₃²⁻ → H₂CO₃

Ba²⁺ + C₆H₁₁NHSO₃⁻ → Ba(C₆H₁₁NHSO₃)₂

The reaction then is:

HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The number of elements must be the same on both sides, so the balanced equation is

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The treatment with H₂SO₄ will produce:

H₂SO₄ → 2H⁺ + SO₄⁻²

Ba(C₆H₁₁NHSO₃)₂ → Ba²⁺ + C₆H₁₁NHSO₃⁻

The balanced reaction will be then:

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃

In the last step, HC₆H₁₁NHSO₃ will react with Ca(OH)₂

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

Ca(OH)₂ → Ca²⁺ + 2OH⁻

The balance reaction will be:

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

3 0

3 years ago

If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?

kodGreya [7K]

Answer:

$[Cu^{2+}]=0.041 M$

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

$[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}$

$[Cu^{2+}]=0.041 M$

Best regards!

6 0

2 years ago

A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the

Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

$\% M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\% =\frac{m_M}{m_M+32.00}*100 \%$