A student sets up a titration apparatus with 25 mL of potassium. hydroxide in the flask. In three to five sentences, explain how the student can use a 0.1 M solution of hydrobromic acid to determine the concentration of potassium hydroxide in the reaction.

Answer:

2,959.36

Explanation:

There are 2959.36 milligrams present in 2.89 x [tex]10^{24}[/tex]  CO2 molecules.

The smallest component of a substance which possesses both its chemical as well as physical characteristics. One or even more atoms make up molecules.

Such a measure of mass as well as weight that weighs 0.0154 grains or one-thousandth of a gram.

Calculation of milligram in CO2 molecule.

Mass = No. of molecule / molar mass.

Molar mass = 44.02 g/mol.

Mass = 2.89 x [tex]10^{24}[/tex]  /[tex]6.023 * 10^{23}[/tex]× 44.02 g/mol.

Mass = 2959.36 milligrams

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Answer:

[Ar] 4s²

Explanation:

The full electron configuration for the Ca+ ion is:

1s² 2s² 2p⁶ 3s² 3p⁶.

To determine the full electron configuration for a Ca+ ion, we need to remove one electron from the neutral calcium atom (Ca). The neutral calcium atom has the electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².

When Ca loses one electron to form the Ca+ ion, the electron removed comes from the highest energy level, which is the 4s orbital.

Therefore, the full electron configuration for the Ca+ ion is:

1s² 2s² 2p⁶ 3s² 3p⁶.

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Using Charles law

[tex]\\ \sf\longmapsto V_1T_2=V_2T_1[/tex]

[tex]\\ \sf\longmapsto V_2=V_1T_2\div T_1[/tex]

[tex]\\ \sf\longmapsto V_2=\dfrac{49.8(356)}{291}[/tex]

[tex]\\ \sf\longmapsto V_2=\dfrac{17728.8}{291}[/tex]

[tex]\\ \sf\longmapsto V_2=60.9mL[/tex]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

[tex]5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4[/tex]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

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The equilibrium constant shows the extent of dissolution of a substance in solution. The molar solubility of the compound here is 1.2 * 10^-8 M.

The solubility product is the equilibrium constant that shows the extent to which a substnace is dissolved in soution.

Now let us set up the ICE table as shown;

               Cr(OH)3 (s) <-> Cr^3+ (aq) + 3OH^- (aq)

I                                           0                 0

C                                          +x               +3x

E                                          x                   3x

Ksp = [Cr^3+] [OH^- ]^3

6.3x10^-31 = [x] [3x]^3

6.3x10^-31 = 27x^4

x = 4√6.3x10^-31 /27

x = 1.2 * 10^-8 M

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The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.

The solubility of a substance in water is in accordance with the solubility rules.  It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.

Now, we will consider each reaction individually to decode whether or not a precipitate is possible.

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#5

At 22min the temperature is at 40°C

#6

Graph ended at 20°C so condetion occured at 20°C

#7

After 70°C the matter got melted do its at liquid state

We want to find the concentration of the resulting solution of mixing 35mL of a given solution and 0.225 L of water.

We will find that the concentration is 13.5%.

First, we need to write both of these in the same units.

We know that:

1000 mL = 1L

Then:

35 mL = 0.035 L

So when we mix the solution and the water we end with a total volume of:

V = 0.035L + 0.225L = 0.260L

The concentration of solution will be equal to the quotient between the volume of solution we used (0.035 L) and the total volume of the mixture (0.260L)

We will get:

Concentration = (0.035L)/(0.260L) = 0.135

Now to write it as a percentage, we only need to multiply this by 100%.

0.135*100% = 13.5%

This means that the mixture has a concentration of the 13.5%.

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Taking into account the definition of density,  the mass of two cubic inches (2.00 in³) of gold is 632.54 grams.

But first you must know the definition of density. Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words,  density is a scalar quantity referring to the amount of mass in a given volume of a substance, that is, it shows the relationship between the weight (mass) of a substance and the volume it occupies.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

[tex]density=\frac{mass}{volume}[/tex]

In this case, you know that:

Replacing in the definition of density:

[tex]19.3\frac{g}{cm^{3} } =\frac{mass}{32.7741 cm^{3} }[/tex]

Solving:

mass= 19.3 [tex]\frac{g}{cm^{3} }[/tex]× 32.7741 cm³

mass= 632.54 g

In summary, the mass of two cubic inches (2.00 in³) of gold is 632.54 grams.

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Cobalt-60 is comprised of 27 protons, 33 neutrons, and 27 electrons. Nuclear reactors are used to artificially create it.

A radioactive isotope containing cobalt with something like a half-life of 5.2713 centuries is called cobalt-60 (60Co). Nuclear reactors are used to artificially create it. Neutron irradiation of bulk material of the monoisotopic as well as mononuclidic cobalt element 59 Co is necessary for deliberate industrial manufacture.

The accidentally generated 60 Co in the latter instance (inside the absence of additional cobalt) is mostly the product of many phases involving neutron activation on iron isotopes inside the reactor. Cobalt-60 is comprised of 27 protons, 33 neutrons, and 27 electrons.

Therefore,  cobalt-60 is comprised of 27 protons, 33 neutrons, and 27 electrons.

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Answer:

Drawing conclusions

Explanation:

The reaction Pb(NO3)2(aq) + NiCl2(aq)---------> PbCl2(s) + Ni(NO3)2(aq) is a precipitation reaction.

A chemical reaction is said to occur when two or more substances are combined to produce new substances. Chemical reactions are classified based on the kind of change taking place in the reaction.

In the reaction;

Pb(NO3)2(aq) + NiCl2(aq)---------> PbCl2(s) + Ni(NO3)2(aq)

We can see that a solid is formed when lead II nitrate reacts with nickel II chloride. This is a precipitation reaction.

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Answer:

27)

a. Silver nitrate

b. Lead chromate

c. Anhydrous magnesium acetate

d. Copper nitrate

28) a. KNo3

Answer:

Seven methods were used to assess water quality in the river system. These methods included single-factor assessment, water quality grading, comprehensive pollution index, the Nemerow pollution index, principle component analysis, fuzzy comprehensive evaluation, and comprehensive water quality identification index.

Answer:

Aluminium sulphate is an inorganic compound.

Answer:

Explanation:

i would say not d im not fully sure but id say c

When molecules present in a bowl of hot soup then they do vibration due to high kinetic energy.

Molecules are the example of chain substance, which can be prepared by the addition of two or more than 2 atoms of same or different nature.

In a hot soup bowl, kinetic energy of the molecules present in that soup increases due to which they move with higher velocity and also shows vibration in their mean position.

Hence, option (C) is correct i.e. vibrate.

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Answer:

Aspartame produces 4 kilocalories of energy per gram when metabolized, sucrose (table sugar) produces 3.9 kilocalories. However, aspartame is approximately 200 times sweeter than sucrose, so it is consumed in much smaller doses.

Explanation:

Answer is Aspartame.

In the pond, the percentage of fish left after ten days is 4%.

Given that;

N(t) = Noe^-kt

Where;

N(t) = percentage of fish at time t

No = Initial percentage of fish

k = rate constant

t = time taken

Given that we had 100% fish population at the beginning, we can see that this reaction is first order.

The rate constant is obtained from;

When t = 5

19 = 100e^-5k

19/100 =e^-5k

0.19 = e^-5k

ln 0.19 = -5k

k = ln (0.19)/-5

k = 0.33

In ten days;

N(10) = 100e^-0.33(10)

N(10) = 4%

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Answer:

htyiu

Explanation:

The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.

The mass in grams of  1.50 x 10²³ atoms of Mg is 5.98 g.

The mass in grams of  4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.

The mass in grams of  8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.

The mass in grams of  25 atoms of W is 3.1 x 10⁻²¹ g.

The mass in grams of  1 atom of Au is 3.27 x 10⁻²² g.

The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;

6.023 x 10²³ atoms = 19 g of F

3.011 x 10²³ atoms F  = ?

= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)

= 9.5 g

The mass in grams of  1.50 x 10²³ atoms of Mg is calculated as follows;

6.023 x 10²³ atoms = 24g of Mg

1.5 x 10²³ atoms F  = ?

= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)

= 5.98 g

The mass in grams of  4.50 x 10¹² atoms of Cl is calculated as follows;

6.023 x 10²³ atoms = 35.5 g of Cl

4.5 x 10²³ atoms Cl  = ?

= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)

= 2.65 x 10⁻¹⁰ g

The mass in grams of  8.42 x 10¹⁸ atoms of Br is calculated as follows;

6.023 x 10²³ atoms = 80 g of Br

8.42 x 10¹⁸ atoms Br = ?

= (8.42 x 10¹⁸  x 80 g)/(6.023 x 10²³)

= 1.12 x 10⁻³ g

The mass in grams of  25 atoms of W is calculated as follows;

6.023 x 10²³ atoms = 74 g of W

25 atoms W = ?

= (25  x 74 g)/(6.023 x 10²³)

= 3.1 x 10⁻²¹ g

The mass in grams of  1 atom of Au is calculated as follows;

6.023 x 10²³ atoms = 197 g of Au

1 atom of Au = ?

= (1  x 197 g)/(6.023 x 10²³)

= 3.27 x 10⁻²² g

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Answer:

The Lewis structure for XeF4 has a total of 36 valence electrons. When we are done adding valence electrons we check each atom to see if it has an octet (full outer shell). We also need to check to make sure we only used the number of available valence electrons we calculated earlier (no more, no less).

Explanation:

Answer:

They want to know how long it has been since it last erupted.

Explanation:

Scientists study a volcano's history to try to predict when it will next erupt.  They also want to know the time span between its previous eruptions

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