How is neutrophilia defined?​

Answers

Answer 1
here i found this website about neutrophils that’s helpful i will link it in the comments

Related Questions

Only animals and not plants can adapt to their environment true or false

Answers

Answer:

true

Explanation:

np have a good day brainliest?

What's a house hold item that has nuclear energy

Answers

Answer:

Electricity, Weapons, Medicine · Food Treatments, ect.

What is the ratio by atoms of elements present in hafnium phosphite?
1:3:4
3:6:2
3:4:12
1:6:15

Answers

3:6:2 is the correct answer

The most common experimental technique to perform elemental analysis is combustion analysis, where a sample is burned in a large excess of oxygen and the combustion products are trapped in a variety of ways. A 99.99% pure, 0.4808 g sample containing only carbon, hydrogen, and nitrogen is subjected to combustion analysis, resulting in the formation of 0.6859 g CO2, 0.6973 g H2O, and 0.4646 g NO. What is the empirical formula of the sample

Answers

Answer:

C₂H₅N₂

Explanation:

In the combustion analysis, all NO comes from the nitrogen of the sample, that means:

Moles NO = Moles N in the sample

In the same way:

Moles CO2 = Moles C in the sample

Moles H2O = 1/2 moles H in the sample

To solve this problem we must convert the mass of each gas to moles in order to find the moles of each atom. The empirical formula is the simplest whole number ratio of atoms presents in the molecule:

Moles NO = N -Molar mass: 30g/mol

0.4646g * (1mol / 30g) = 0.01549 moles N

Moles CO2 = C -Molar mass: 44.01g/mol-

0.6859g * (1mol / 44.01g) = 0.01559 moles C

Moles H2O = 1/2moles H -Molar mass: 18.02g/mol-

0.6973g * (1mol / 18.02g) = 0.03870 moles H

The ratio of atoms is -Dividing in the low number of moles = moles N-:

C = 0.01559 / 0.01549 = 1

N = 0.01549 / 0.01549 = 1

H = 0.03870 / 0.01549 = 2.5

Twice this ratio of atoms -Because empirical formula must be given only with whole numbers:

C = 2

N = 2

H = 5

The empirical formula is:

C₂H₅N₂

Part A
How many moles of chlorine gas are needed to make 0.6 moles of sodium chloride?
Given the reaction: 2Na + Cl2 + 2NaCl
O 1.2
O 0.6
0 3.6
O 0.3
not enough information
Submit
Request Answer

Answers

Answer:

[tex]n_{Cl_2}=0.3molCl_2[/tex]

Explanation:

Hello there!

In this case, according to the given chemical reaction whereas the sodium chloride is in a 2:1 mole ratio with chlorine, the required moles of the later are computed as shown below:

[tex]n_{Cl_2}=0.6molNaCl*\frac{1molCl_2}{2molNaCl}[/tex]

So we cancel out the moles of NaCl to obtain:

[tex]n_{Cl_2}=0.3molCl_2[/tex]

Best regards!

Identify the term that matches each electrochemistry definition.

a. The electrode where reduction occurs ___________
b. An electrochemical cell powered by a spontaneous redox reaction ___________
c. The electrode where oxidation occurs__________
d. An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction ____________
e. A chemical equation showing either oxidation or reduction ___________

Answers

Answer: a. Cathode

b. Galvanic cell

c. Anode

d. Electrolytic cell

e. half reaction

Explanation:

Galvanic cell or Electrochemical cell is defined as a device which is used for the conversion of the chemical energy produced in a spontaneous redox reaction into the electrical energy.

Electrolytic cell is a device where electrical energy is used to drive a non spontaneous chemical reaction.

In the electrochemical cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.  Thus the electrons are produced at anode and travel towards cathode.

The balanced two-half reactions will be:

Oxidation half reaction : [tex]M\rightarrow M^{n+}+ne^-[/tex]

Reduction half reaction : [tex]N^{n+}+ne^-\rightarrow N[/tex]

Thus the overall reaction will be: [tex]M+N^{n+}\rghtarrow M^{n+}+N[/tex]

Help (also tell me what the phenotype and genotypes are)

Answers

Answer:

genotype are the organism's hereditary information for example DNA

phenotype are the organism's physical characteristics for example your eyes or hair

or how your nose looks

Imine formation from an aldehyde and an amine proceeds reversibly under slightly acidic conditions. The reaction is reversible due to acid-catalyzed hydrolysis of the imine. Imine formation generates an equivalent of water. Given these facts, explain why the imine can be isolated from the reaction mixture.

Answers

Answer:

Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber

Explanation:

In this reaction, a non -aqueous solvent  is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.

If the number of moles of a gas doubles, its volume will also double
according to
*
Please hurry! What is the answer to this?

Answers

Explanation:

Boyle's Law

2. A plant growing in response to the sunlight.
A.Gravitropism
B.Phototropism
C.Thigmotropism
D.Hydrotropism

Answers

Answer:

phototropism

Explanation:

since photo is (light?

Phototropism - because she photo means light hydro is water and gravi thing of earth

If I have 7.0 x 10^24 formula units of magnesium chloride, how many grams of chlorine would I have?

Answers

Answer:

824 g

Explanation:

First we convert 7.0 x 10²⁴ formula units of magnesium chloride (MgCl₂) into moles, using Avogadro's number:

7.0 x 10²⁴ formula units ÷ 6.023x10²³ formula units/mol = 11.62 mol MgCl₂

There are two Cl moles per MgCl₂ moles, meaning that we have (2 * 11.62) 23.24 moles of Cl

Finally we convert 23.24 moles of chlorine into grams, using chlorine's molar mass:

23.24 mol * 35.45 g/mol = 824 g

Draw a structural formula for the organic product formed by treating butanal with the following reagent: NaBH4 in CH3OH/H2O You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Do not draw organic or inorganic by-products.

Answers

Answer:

Please find the solution in the attachment file.

Explanation:

How many grams of CuSO4 are required to make a 5.0 ml solution that has a concentration of 400 mg/ml?

Answers

Answer:  molecular weight of CUSO4 or grams. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CUSO4, or 346.10221 grams. Note that rounding errors may occur, so always check the results.

Explanation:

3. Convert the word equations below to symbolic:
Sodium + chlorine sodium chloride
• Calcium + bromine ---- calcium bromide
• Potassium + water → potassium hydroxide + hydrogen
ot

Answers

Answer:

1)2Na + Cl2 ----> 2NaCl

2)Ca + Br2 ---->CaBr2

3)K + H2O -----> KOH + H2

Helen is studying the properties of solids. She is comparing the wooden cube and the steel cube shown in the picture below.

mc065-1

Which property is the same for the two cubes?


color

luster

texture

volume

Answers

volume. they are different colors, they aren’t the same luster, and one is smooth while the other might be slightly rough

How many grams of hydroxide pellets, NaOh are required to prepare 50.0ml of a 0.150 M solution ?

A.0.300g
B.3.00g
C.2.00g
D.200.g
E. 0.025 M

Answers

the correct answer is E10.
(O.0500L) x (0.150mo/L)x(39.997 15 g NaOh/mol)=0.300 g NaOh



ANSWER A

Express each of the following answers in the IUPAC format. Do not include any capitals or spaces in your name. Separate multiple substituent numbers using a comma. Use a hyphen between numbers and names. (ex: 1,2-dichloro-3-methylcyclopentane). Part A Spell out the full name of the molecule. Part B Spell out the full name of the molecule. Part C Spell out the full name of the molecule. Part D Spell out the full name of the molecule.

Answers

The full question is shown in the image attached

Answer:

See explanation

Explanation:

In naming an alkane, the first thing we do is to obtain the parent chain by counting the number of carbon atoms in the chain.

When we obtain that, then we identify the substituents and number them in such a way that they have the lowest numbers. The compounds shown have the following names according to the order in which the structures appear in the image attached;

1. 2-methyl propane

2. 2,4-dimethyl heptane

3. 2,2,3,3-tetramethyl butane

4. 5-ethyl-2,4-dimethyl octane

the experimental control is the ?
A. Dependent variable
B. Design of the experiment.
C. Outcome of the experiment
D. the standard against which the results are measured .

Answers

Answer:

D. In an experiment, one of the variables will be giving a placebo and the other, a treatment

Explanation:

Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.

A 1.4639-g sample of limestone is analyzed for Fe, Ca, and Mg. The iron is determined as Fe2O3 yielding 0.0357 g. Calcium is isolated as CaSO4, yielding a precipitate of 1.4058 g, and Mg is isolated as 0.0672 g of Mg2P2O7. Report the amount of Fe, Ca, and Mg in the limestone sample as %w/w Fe2O3, %w/w CaO, and %w/w MgO

Answers

Answer:

A) w/w % of Fe - 2.44%

B) w/w % of Mg - 4.590%

C) w/w % of Ca - 96%

Explanation:

A) w/w % of Fe in limestone as Fe2O3 = (Mass of Fe2O3 /Mass of limestone) x 100

 0.0357/1.4639 X 100

= 2.438 =2.44

B) w/w % of Mg in limestone as Mg2P2O7 = (Mass of Mg2P2O7 /Mass of limestone) x 100

= 0.0672/1.4639 X 100

= 4.590

C) w/w % of Ca in limestone as CaSO4 = (Mass of CaSO4/Mass of limestone) x 100

= 1.4058/1.4639 X 100

= 96

                                                           

Answer:Fe =1.71%    Ma=0.99%    Ca=28.24%

Explanation:

Identify the Lewis base in this balanced equation: 6 upper H subscript 2 upper O plus UPper C r (upper H subscript 2 upper O) subscript 6 superscript 3 plus. H2O Cr3+ Cr(H2O)63+

Answers

The Lewis base that we can see in the equation is [tex]H_{2} O[/tex]

What is the Lewis base?

We have to bear in mind that the Lewis base is the substance that is able to donate a pair of electrons. The substances that is able to accept the pair of electrons is what we call the Lewis acid.

The Lewis acid must possess a space for the addition of an electron pair while the Lewis base would have to be a specie that has a lone pair of electrons or has a negative charge. Thus the water molecule would serve as a Lewis base in this regard.

Learn more about Lewis base;https://brainly.com/question/15570523

#SPJ1

Answer:

6 upper H subscript 2 upper O plus UPper C r (upper H subscript 2 upper O) subscript 6 superscript 3 plus.

✔ H2O or A.)

Cr3+

Cr(H2O)63+

select two correct answers

Answers

Answer:

D and E

Explanation:

The other answers don't support steel, they support iron or are supporting both, and the question is why steel alloys are more often used

C supports both

B supports how iron's strength

A supports iron being easily shaped and bent

hope this helps

- Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCI (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCI(g) → H2(g) + Cl2(g)
AH = +184.6 KJ
2H2(g) + 1/2 N2(g) + 1/2 Cl2(g) → NH4Cl(s) deltaH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
deltaH = +184.6 kJ​

Answers

Answer:

Explanation:

We have the three equations:

[tex]NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)} ..... \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)} => 2NH_{3(g)} .... \Delta H = +184.6kJ (4)[/tex]

(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.

In another experiment, the student titrated 50.0 mL of 0.100 M HC,H,O, with
0.100 M NaOH(aq). Calculate the pH of the solution at the equivalence point

Answers

Answer:

Eqv Pt pH = 8.73

Explanation:

    HOAc                   +            NaOH            =>            NaOAc              + H₂O

50ml(0.10M HOAc)  +  50ml(0.10M NaOH) => 100ml(0.05M NaOAc) + H₂O

For neutralized system, 100ml of 0.05M NaOAc remains

NaOAc => Na⁺ + OAc⁻

Na⁺ + H₂O => No Rxn

          OAc⁻  +  H₂O  => HOAc + OH⁻

C(i)   0.05M       -----        0M      0M

ΔC        -x           -----         +x        +x

C(f)    0.05-x      

       ≅ 0.05M    -----          x          x

Kb = Kw/Ka = [HOAc][OH⁻]/[OAc⁻] = 1 X 10⁻¹⁴/1.7 X 10⁻⁵ = (x)(x)/(0.05M)

=> x = [OH⁻] = SqrRt(0.05 x 10⁻¹⁴/1.7 x 10⁻⁵) = 5.42 x 10⁻⁶M

=> pOH = -log[OH⁻] = -log(5.42 x 10⁻⁶) = 5.27

pH + pOH = 14 => pH = 14 - pOH = 14 - 5.27 = 8.73 Eqv Pt pH

please answer asap!

What is the molarity of a KOH solution if 200 ml of the solution contains 0.6 moles KOH?

a. 0.3 M
b. 0.6 M
c. 3.0 M
d. 6.0 M

Answers

200 ml is 1/5 of a liter, so the answer is five times the number of moles present in the solution. 0.6 moles/0.2 liter = x moles/1.0 liter. Solving for x gives 0.2 x = 0.6 or x = 3.0 M

so the answer is c

I will mark brainliest
What kind of weather forms with an occluded front?

Answers

Answer:Occluded fronts usually form around areas of low atmospheric pressure. There is often precipitation along an occluded front from cumulonimbus or nimbostratus clouds. Wind changes direction as the front passes and the temperature either warms or cools.

Explanation:

Occluded fronts usually form when the area is of low atmospheric pressure.

A chemist titrates 0.200 M NaOH, strong base, with 50.00 ML of 0.150 M HCI, strong acid. How many mL of NaOH will be required to titrate to the endpoint

Answers

[tex](normality \: of \: acid)×(volum \: of \: acid) = (normality \: of \: base)×(volum \: of \: base)[/tex]

0.15N × 50mL = 0.2N × (Vbase)

75mL = Volum of base

37.5mL of NaOH will be required to titrate 0.200 M NaOH, strong base, with 50.00 ML of 0.150 M HCI, strong acid to the endpoint.

How to calculate volume?

The volume of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

C1 = initial concentrationC2 = final concentrationV1 = initial volumeV2 = final volume

C1 = 0.200MC2 = 0.150MV1 = ?V2 = 50mL

0.2 × V1 = 0.150 × 50

0.2V1 = 7.5

V1 = 7.5/0.2

V1 = 37.5mL

Therefore, 37.5mL of NaOH will be required to titrate 0.200 M NaOH, strong base, with 50.00 mL of 0.150 M HCI, strong acid to the endpoint.

Learn more about volume at: https://brainly.com/question/26416088

Calculate the pH of a solution with [H+] = 1.8 x 10^-5 M

Answers

Answer:

4.74

Explanation:

pH= -log[H+]

pH= -log(1.8 x 10^-5)= 4.74

Because a water molecule has a negative end and a positive end it displays?

Answers

Answer:

A water molecule displays polarity by having negative and positive ends. Due to this charge difference, a water molecule is called a dipole.

Explanation:

What is the molarity of a 9.79 L solution that contains 0.400 mol HCI

Answers

Answer:

0.041 M

Explanation:

The molarity of a solution when given the number of moles the substance and it's volume can be found by using the formula

[tex]c = \frac{n}{v} \\ [/tex]

n is the number of moles

v is the volume in L

We have

[tex]c = \frac{0.4}{9.79} \\ = 0.04058...[/tex]

We have the final answer as

0.041 M

Hope this helps you

What is the fundamental different between combustion and nuclear reaction

Answers

Answer:

Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclear.

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