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Soloha48 [4]
3 years ago
14

Someone please help me !!

Chemistry
1 answer:
Lisa [10]3 years ago
4 0

Answer:

the organism reproduces asexually

Explanation:

it only needs itself and it uses cell fission

You might be interested in
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
Calculate the Ph and the POH of an aqueous solution that is 0. 040 m in HCl(aq) and 0. 075 m in HBr(aq) at 25 °C.
LenaWriter [7]

pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.

pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.

pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.

Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.

Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :

HCL(0.040M)----> H+(0.040M) +CL-(0.040M)

HBr(0.075M)----> H+(0.075M) +Br-(0.075M)

so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.

pH is calculated as:

pH = -log[H+]

Substituting values in the equation:

log(0.115M)= 0.939 pH

pOH is calculated as:

14 - pH = pOH

Substituting values in the equation above:

14 - 0.939= 13.061 pOH

Therefore, pH is 0.939 and pOH is 13.061.

Learn more about pH and pOH here:

brainly.com/question/2947041

#SPJ4

5 0
2 years ago
Assuming equal concentrations, arrange these solutions by pH.
LenKa [72]

Answer:

Highest pH(most basic)

Sr(OH)2(aq)

KOH (aq)  

NH3(aq)

HF (aq)  

HClO4(aq)

Lowest pH(most acidic)

Explanation:

The concentration of H+ ion will determine the pH of a solution. The pH actually reflects the ratio of H+ ion and OH- since both of them can combine into water. Solution with more H+ ion will have a lower pH and called acidic, while more OH- will have high pH and be called basic. Strong acid/base will be ionized more than weak acid/base.

Sr(OH)2(aq) = strong base, release 2 OH- ion per mole

KOH (aq) = Strong base, release 1 OH- per mole

NH3(aq) = weak base, release less than 1 OH- per mole

HF (aq) =strong acid, release 1 H+ per mole

HClO4(aq) = stronger acid, release 1 H+ per mole

8 0
3 years ago
A calorimeter contains 22.0 mL of water at 14.0 ∘C . When 2.50 g of X (a substance with a molar mass of 82.0 g/mol ) is added, i
Alik [6]

Answer:

The enthalpy change in the the reaction is -47.014 kJ/mol.

Explanation:

X(s)+H_2O(l)\rightarrow X(aq)

Volume of water in calorimeter = 22.0 mL

Density of water = 1.00 g/mL

Mass of the water in calorimeter = m

m=1.00 g/mL\times 22.0 mL=22 g

Mass of substance X = 2.50 g

Mass of the solution = M = 2.50 g + 22 g = 24.50 g

Heat released during the reaction be Q

Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C

Specific heat of the solution is equal to that of water :

c = 4.18J/(g°C)

Q=Mc\times \Delta T

Q=24.50 g\times 4.18 J/g ^oC\times 14.0^oC=1,433.74 J=1.433 kJ

Heat released during the reaction is equal to the heat absorbed by the water or solution.

Heat released during the reaction =-1.433 kJ

Moles of substance X= \frac{2.50 g}{82.0 g/mol}=0.03048 mol

The enthalpy change, ΔH, for this reaction per mole of X:

\Delta H=\frac{-1.433 kJ}{0.03048 mol}=-47.014 kJ/mol

5 0
3 years ago
What is the empirical formula for a compound if a sample contains 1. 0 g of S and 1. 5 g of O? SO SO3 S2O2 S2O3.
Tcecarenko [31]

The empirical formula of the given compound is \bold{SO_3}.

The correct option is B.

<h3>What is an empirical formula?</h3>

The empirical formula of a chemical compound is the simplest whole-number ratio of atoms contained in the substance.

Given,

1.0 g of S

1.5 g of O

To calculate the empirical formula, we will divide the masses of the elements by their atomic weight.

For sulfur

\bold{\dfrac{1.0}{32} =0.03125\; mol}

For oxygen

\bold{\dfrac{1.5}{16} =0.09375\; mol}

Now, divide the greater value of mole came by the smaller value

\bold{\dfrac{0.09375\; mol}{0.03125\;mol} = 3}

Thus, the empirical formula for the given compound is 1 for S and 3 for O

\bold{SO_3}

Learn more about empirical formula, here:

brainly.com/question/11588623

7 0
2 years ago
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