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yan [13]
2 years ago
15

Is utensils a substance homogeneous mixture or heterogeneous mixture

Chemistry
1 answer:
Phoenix [80]2 years ago
7 0

A pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.

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A curium-245 nucleus is hit with a neutron and changes as shown by the equation. Complete the equation by filling in the missing
Lera25 [3.4K]

Try this option:

^{245}_{96}Cm+^1_0n=^{103}_{42}Mo+^{245+1-103-3*1}_{96+0-42-3*0}(X)+3^1_0n;

^{245}_{96}Cm+^1_0n=> ^{103}_{42}Mo+^{140}_{54}(X)+3^1_0n;

^{245}_{96}Cm+^1_0n=> ^{103}_{42}Mo+^{140}_{54}Xe+3^1_0n;

Answer:

^{140}_{54}Xe

8 0
2 years ago
Read 2 more answers
Consider the following reaction (X = Cl or Br) which statement s is are correct?
statuscvo [17]

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

5 0
2 years ago
6.An acid-base indicator is usually a weak acid with a characteristic color in the protonated and deprotonated forms. Because br
Eddi Din [679]

Answer:

At equal concentration of HBCG and BCG^-, the colour is green. This colour first appears at pH = 3.8

Explanation:

HBCG is an indicator that is prepared by dissolving the solid in ethanol.

Since

Ka=[BCG−][H3O+][HBCG]When [BCG-] = [HBCG], then Ka = [H3O+].

If pH = 3.8

Ka= [H3O+] = -antilog pH = -antilog (3.8)

Ka= 1.58 ×10^-4

5 0
2 years ago
What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi
kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s).

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq).

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

How to rewrite a chemical equation to produce a net ionic equation?

  1. Rewrite all reactants and products that ionizes completely in the solution as ions.
  2. Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

  • Soluble salts,
  • Strong acids, and
  • Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

  • MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as \rm Mg^{2+} and \rm 2\; Cl^{-}.
  • NaOH is a strong base. Each unit of NaOH can be written as \rm Na^{+} and \rm OH^{-}.
  • Mg(OH)₂ is a weak base and should not be written.
  • NaCl is a salt and is soluble in water. Each unit of NaCl can be written as \rm Na^{+} and \rm Cl^{-}.

\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}.

Ions on both sides of the equation:

  • \rm 2\; Cl^{-}, and
  • \rm 2\; Na^{+}.

Add the state symbols:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

For the second reaction:

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}.

\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s).

5 0
2 years ago
Read 2 more answers
An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an
timofeeve [1]

Answer:

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance: M_{1} = 175 + 11M_{2}

Water: 0.804M_{1} = 175 + 0.6M_{2}

Using substitution method, we have:

M_{1} = 220.77 kg/s

M_{2}  = 4.16 kg/s

The maximum production rate of solid crystal is 10M_{2} = 10*4.16 = 416 kg/s

Around evaporator:

0.45M_{5} = 175

M_{5} = 175/0.45 = 389 kg/s

Around the mixing point:

M_{1} + M_{3} = M_{4} + M_{5}

Solid crystal: 0.196M_{1} + 0.4M_{3} = M_{4}

Using the last two equations, we can obtain:

0.804M_{1} + 0.6M_{3} = M_{5}

0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5

M_{3} = 211.5/0.6 = 352.5 kg/s

The rate of supplying fresh feed to obtain the production rate is:

\frac{M_{3}}{M_{1}} = 352.5/220.77 = 1.6

7 0
3 years ago
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