A 2.0% (w/v) solution of sodium hydrogen citrate, Na2C6H6O7, which also contains 2.5% (w/v) of dextrose, C6H12O6, is used as an anticoagulant for blood which is to be used for transfusions. What is the molarity of the sodium hydrogen citrate in the solution

Answer:

$4.49 g/cm3

Explanation:

Density=Mass/Volume

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Answer: 1.2 x 10^-17 microounces

Explanation:

Ounce = 28.5G microounce = 28.5*10^-6g

3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17

Explanation:

H2SO3 is more acid than H2TeO3. Since S is more electronegative than Te is. In H2SO3, thus, dissociation of H+ would be smoother.

So, H2SO3's got high Ka.

HCH3SO2 is more acid than HCH3CO2. Since S is more electronegative than C. So, HCH3SO2 is a high Ka.

HClO2 is more acid than HClO. Since in HClO2, after the donation of H+ ion, the negative charge is set by two oxygen atoms, while in HClO, only one oxygen atom stabilizes the negative charge.

So, HClO2 is a high Ka

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

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Answer:

The correct option is d. ion exchange chromatography and gel filtration chromatography

Explanation:

The mixture contains proteins with different isoelectric points and molecular weights. Ion exchange chromatography separates the proteins with different electric charges, which depend on the isoelectric point. At certain pH of the aqueous medium, the proteins will have a certain electric charge and they will be retained differently in the stationary phase. Proteins A and D have similar isoelectric points (around 4.0) and maybe they cannot be efficiently separated by this method; the same occurs with protein B and C (with an isoelectric point around 9.0). To separate proteins A from D and protein B from C, it could be used gel filtration chromatography. This separation method is based on differences in molecular weights. So, at the end of the process, we will obtain the proteins A, B, C, D separated one from each other.

Answer:

d. 9.95 × 10⁻³ mol

e. 249 g

f. 2.55 × 10²⁴ atoms

g. 4.32 × 10²² atoms

Explanation:

d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol

e. the mass of 4.23 mol of cobalt

The molar mass of cobalt is 58.93 g/mol.

4.23 mol × 58.93 g/mol = 249 g

f. the number of cobalt atoms in 4.23 mol of cobalt

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms

g. the number of cobalt atoms in 4.23 g of cobalt

First, we will calculate the moles of cobalt using the molar mass of cobalt.

4.23 g × 1 mol/58.93 g = 0.0718 mol

Then, we will calculate the number of cobalt atoms using Avogadro's number.

0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

   So;

    0.57mole of hydrogen gas will also produce 0.57 water

Answer:

Explanation:

Nitric oxide is the gas NO, it reacts with oxygen as shown below;

2NO(g) + O2(g) -----> 2NO2(g)

Now the gas formed is the gas NO2 which is known to be reddish brown in colour.

A diagrammatic representation of this reaction is shown in the image attached to this answer.

Image credit: Chemlibretext

Answer:bro u already have an answer why are you asking?

Explanation:

Answer:

A. This allows the plants to conserve water and not bloom during the heat of the day

Explanation:

Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.

Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.

Answer:

A is the answer the reason why is because I got it right when I was doing my test

Answer:

0.696 atoms of oxygen

Explanation:

We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:

Mass of CaWO₄ = 50 g

Molar mass of CaWO₄ = 40 + 184 + (4×16)

= 40 + 184 + 64

= 288 g/mol

Mole of CaWO₄ =?

Mole = mass / Molar mass

Mole of CaWO₄ = 50 / 288

Mole of CaWO₄ = 0.174 mole

Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:

1 mole of CaWO₄ contains 4 atoms of oxygen.

Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.

Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.

Answer:

D

Explanation:

Answer:

D: Electrical -> mechanical

Explanation:

Hope this helps!

Answer:

movement of charged particles through the wire .

Explanation:

When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .

Answer:

Q= Magnesium R= Chlorine

Explanation:

The element Q should be magnesium and R is chlorine.

An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.

The electrostatic attraction leads to the formation of the bond.

 To solve this problem, the hypothetical compound is QR₂

       Mg                        Cl

     2 8 2                    2 8 7

So, Mg transfers 2 electrons to two atoms of chlorine.

 This leads to the formation of the compound MgCl₂

Answer:

In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M

Explanation:

In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10

In the first rinse the concentration must be of 0.9M  10 = 0.09M

2nd = 0.009M

3rd = 0.0009M

4th = 0.00009M

5th = 0.000009M →

Answer:

1.6734 g\ml..hope it helps

Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.

Answer: The enthalpy change for this reaction is, -362.8 kJ

Explanation:

The balanced chemical reaction is,

[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]

[tex]\Delta H=-362.8kJ[/tex]

Therefore, the enthalpy change for this reaction is, -362.8 kJ

Answer:

In advance

middle

lower

Explanation:

These are the safety precautions needed when carrying out duties in the fume hood.

When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal in advance to reduce and minimize the raising and lowering of the hood sash at intervals.

It is also pertinent to understand that working in the middle of the work surface helps to promote the movement of air and keeps the area neat and tidy.

However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to lower the sash at that point in time.

Answer:

2.9moles of hydrogen gas

Explanation:

convert liters to dm³

since 1liter= 1dm³

thus, 65.0liters = 65.0dm³

number of moles = volume given/22.4dm³

= 65.0/22.4

=2.9moles

Answer:

3

Explanation:

For numbers with decimals, count the number after the decimal.

Answer:

hello your question is incomplete attached below is the complete question

answer :

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

Explanation:

Reason for the mechanism

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

attached below is the detailed mechanism

Answer:

See explanation

Explanation:

The atoms of metals have fewer valence electrons than the atoms of metals and metalloids.

Atoms of metals have only very few valence electrons in their outermost shells hence they donate electrons during bonding. However, atoms of nonmetals have more electrons in their outermost shells and rather accept electrons during bonding. The atoms of metalloids just have a number of valence electrons that are intermediate between those of metals and nonmetals and mostly share electrons in covalent bonds.

Similarly, atoms of metallic elements differ from each other in the number of valence electrons present in the valence shell of the atom of each element. For instance, sodium has one electron in the valence shell of its atom while aluminium has three electrons in the valence shell of its atom.

The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and the its bonding pattern.

The atoms of different metals varies in it ability to bond quickly.

The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and how it bonds.

Metallic atoms have very few electrons in the outermost shell. The valency electrons of this metallic atoms are few and are easily lost during bonding. They have the ability to release there valency electrons easily. Example of this metals are sodium, potassium , calcium etc.

On the other hand non metallic elements have numerous electron in the outermost shell and easily receive electron during bonding. Example are chlorine, fluorine, oxygen etc.

The metalloid atoms like silicon and germanium have an average number of electron in their outermost shell. They are in between.

The atoms of different metals varies in it ability to bond quickly. For example the group 1 metals are very reactive than the group 2 metals. This simply means the group 1 metals(alkali metals) goes into bonding more easily than the group 2 metals(alkali earth metals).    

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Answer is given below

Explanation:

given data

weight = 6N

temp = 105° C

weight reduced = 5N

solution

weight of container is 1N

SO W = (6-1) = 5

And Wd = 5 - 1 = 4

so

moisture content is

moisture content = [tex]\frac{W-Wd}{Wd} \times 100[/tex]       .......1

moisture content = [tex]\frac{5-4}{4} \times 100[/tex]

moisture content = 25%

and

as we know density of soil soild = 2700 kg/m³

density of water = 1000 kg/m³

and sp gravity of soil =  [tex]\frac{2700}{1000}[/tex]  = 2.7

so

now we get here bulk unit weight

bulk unit wt = [tex]Yw \times [\frac{G+e}{1+e}][/tex]     ..........2

bulk unit wt = [tex]9.01 \times [\frac{2.7 + 0.675}{1+0.675}][/tex]

bulk unit wt = 19.766 KN/m³

and

so dry unit wt  will be

dry unit wt = [tex]\frac{Ysat}{1+w}[/tex]           ..............3

dry unit wt = [tex]\frac{19.766}{1+0.25}[/tex]

dry unit wt = 15.813 kN/m³

Answer:

I don't get it is it even a question?

1s22s22p63s1. That's the ans