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muminat
2 years ago
11

How to balance this equation

Chemistry
2 answers:
Svetlanka [38]2 years ago
7 0

Answer:

2Al + 3Ni(SO4) ------->   1Al2(SO4)3 +  2Ni

Explanation:

Afina-wow [57]2 years ago
7 0
2Al+3Ni(SO4)-> 1Al2(SO4)3+3Ni
You might be interested in
I've a question from chemistry...
serious [3.7K]
<h3>Answer:</h3>

22.06 Hours

<h3>Explanation:</h3>
  • Half life is the time it takes for a radioactive material or isotope to decay by half of its initial amount.

We are given;

Radioactive isotope Iodine-133

Initial counts per minute = 3150 counts

Counts after decay = 3055 per minute

Time taken = 1 hour

But, we know that to calculate the remaining amount of the radioactive isotope, then,

Remaining amount = Initial amount × 0.5^n , where n is the number of half-lives.

Therefore;

3055 counts = 3150 × 0.5^n

0.5^n = 0.9698

To get n

n = log 1 ÷ log 0.9698

 = 0.044

But, n is given by dividing time and the number of half lifes

Therefore, half life = time ÷ Number of half lives

                              = 1 hour ÷ 0.044

                              = 22.60 hours

Therefore, the half life of iodine-133 is 22.60 hours

3 0
2 years ago
What is indicated by Benedict's solution changing from blue to orange when heated?
Sidana [21]

1.Benedict�s reagent (blue in color):

Benedict's reagent (also called Benedict's solution or Benedict's test) is a reagent named after an American chemist, Stanley Rossiter Benedict. It can be prepared from sodium carbonate, sodium citrate and copper(II) sulfate.

Benedict's reagent is used as a test for the presence of reducing sugars such as glucose, fructose, galactose, lactose and maltose.

In the presence of reducing sugars and heat in water benedict�s reagent will change color from blue to orange/green and sometimes as well orange red.

Benedict's reagent contains blue copper(II) sulfate (CuSO4) which is reduced (takes up lost electrons from the reducing sugar) to a, ORANGE/RED colored red copper(I) oxide (Cu2O) by aldehydes, also oxidizing them to carboxylic acids. The copper(I) oxide is insoluble in water and so precipitates. (wiki sourced).

Summary: benedicts reagent color changes from a blue, to then a green, then final orange red color when electrons from an �unknown� reducing sugar, in a warm water bath suspension.

CLUE: *Benedicts reagent changes color to a Orange/Red color if a reducing sugar is your unknown.

2. Biuret reagent (blue in color normally, unreacted): is made of potassium hydroxide (KOH) and copper (II) sulfate (CuSO4), together with potassium sodium tartrate.

The Biuret blue color (no heat needed) reagent turns Violet in the presence of proteins (also called by some polypeptide chains), and changes to Violet/Pink when combined with short-chain (not long) polypeptides (amin acids {monomers of a protein} linke dtogethe rin a chain by bonds identified as peptide bonds, hence �poly (many) peptides�.

The reagent, Biuret protein assay, is a colorimetric (color change) yest used to determine an unknown solution�s protein concentration.

3.Sudan IV (dark red/burgundy color) is a lysochrome (fat-soluble dye) diazo dye used for staining of lipids, triglycerides and lipoproteins on frozen paraffin sections.

In industry, Sudan IV is used to color nonpolar substances like oils, fats, waxes, greases, various hydrocarbon products, and acrylic emulsions. (wiki sourced). In other words lipids �pick up� the color of the Sudan IV dye.

Scenario:

In a lab room, one (you) is handed three small reagent containers, some beakers, some test tubes, safety gloves and goggles, and SIX unknown solutions in testubes, and a known control of distilled water, all to be tested, by you, with the reagents, for the presence of either MONOSACCHARIDES (reducing sugar varieties), PROTEINS (long or short chain polypeptides) and potential LIPIDS: oils fats and waxes

The six unknowns and distilled water are labeled one, two, three, four, five, six, and seven.

ALL are all clear in color.     Only one of the test tube containers contains pure distilled water.

TEST ONE:

When one uses Benedict�s reagent on samples from each test tube, in a hot water bath heat source: samples ONE and FOUR turn color from Benedict�s blue being added to them, in color, from a blue to green to red-orange color (a color change definitely happens).

Samples TWO, THREE, FIVE, SIX, and, SEVEN change from clear to the Benedict�s reagent blue color when the benedicts was added. In essence NO color change happens with these unknon samples, even in the warm water bath. ONLY ONE and FOUR are positive color changes.

TEST TWO:

When one uses the Biuret reagent on samples from test tubes TWO, and THREE, a Pink color forms, and in sample FIVE the chemical unknown changes to a Violet purple color.

IN Samples   ONE FOUR SIX and SEVEN stay the blue color of the Biuret reagent that was added to the initially clear samples. NO color change occurs in these

TEST THREE:

When one adds Sudan IV to all seven samples ONLY sample SIX reacts and the liquid of the unknown sample becomes a reddish / burgundy color.

Samples ONE, TWO, THREE, FOUR, FIVE, and SEVEN turn color to a dilute red color from the stain added to the clear samples, but not in the intensity that is sample unknown SIX.

What can you deduce for the preceding set of experiments as to what was in each sample regarding: Monosaccharide / reducing sugar, a short chain polypeptide, a long chain polypeptide, or a lipid.  Now you just have to do the samples .

3 0
3 years ago
Read 2 more answers
What is the mass percent concentration of a saline solution prepared by dissolving 1.00 mol of Na2SO4 in 1.00 L of water?
Iteru [2.4K]

Answer:

good luck hope you do good

Explanation:

good luck

8 0
3 years ago
I need help on this science question
ICE Princess25 [194]

Answer:

C

Explanation:

The kinetic energy of a body is the energy due to the motion of a body. It is mathematically given as;

 Kinetic energy  = \frac{1}{2} m v²

m is the mass of the body

v is the velocity

 Now, let us find the K.E from the given choices;

Option C will have the greatest kinetic energy;

         mass  = 0.14kg , v  = 40m/s

 Kinetic energy  = \frac{1}{2} x 0.14 x 40²   = 112J

5 0
3 years ago
Determine the pH at the point in the titration of 40.0 mL of 0.200 M H₂NNH₂ with 0.100 M HNO₃ after 100.0 mL of the strong acid
Irina-Kira [14]

Answer:

pH = 1.85

Explanation:

The reaction of H₂NNH₂ with HNO₃ is::

H₂NNH₂ + HNO₃ → H₂NNH₃⁺ + NO₃⁻

Moles of H₂NNH₂ and HNO₃ are:

H₂NNH₂: 0.0400L ₓ (0.200mol / L) = 8.00x10⁻³ moles of H₂NNH₂

HNO₃: 0.1000L ₓ (0.100mol / L) = 0.01 moles of HNO₃

As moles of HNO₃ > moles of H₂NNH₂, all H₂NNH₂ will react producing H₂NNH₃⁺, but you will have an excess of HNO₃ (Strong acid).

Moles of HNO₃ in excess are:

0.01 mol - 8.00x10⁻³ moles = 2.00x10⁻³ moles of HNO₃ = moles of H⁺

Total volume is 100.0mL + 40.0mL = 140.0mL = 0.1400L.

Thus, [H⁺] is:

[H⁺] = 2.00x10⁻³ moles / 0.1400L = 0.0143M

As pH = - log [H⁺]

<h3>pH = 1.85 </h3>
6 0
2 years ago
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