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3 years ago

What is a picture of homologous pairs of chromosomes and sex chromosomes?

Chemistry

2 answers:

Alecsey [184]3 years ago

5 0

Answer:

allene?

Explanation:

ANTONII [103]3 years ago

4 0

Allele

Hope this helps :)

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What is the name of Na

amid [387]

Sodium
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8 0

2 years ago

Please help me do three

Orlov [11]

Answer:

3a) (NH₄)₂ CO₂

b) FeF₂

c) CaN₂

d) Al₂O₃

e) NiCl₂

f) Mg₃P₂

g) KHCO₃

h) PbOH₂

6 0

2 years ago

Twenty-seven milliliters of an acid with an unknown concentration are titrated with a base that has a concentration of 0.55 M. T

Svetllana [295]

It should be 0.25 M. Use the formula C1*V1=C2*V2, for those values, as it is right when it changed colour. Remember to change the if those are not the same (but in your case it is, so no need this time).

C1*V1=C2*V2
C1*27ml=0.55M*12.5ml
C1=(0.55M*12.5ml)/27ml = 0.25M

4 0

3 years ago

Read 2 more answers

The heat of vaporization of 1–pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K•mol. What is the approximate b

tensa zangetsu [6.8K]

Answer:

375 K

Explanation:

Using the experssion shown below as:

$\Delta G^0=\Delta H^0_{vap}-T\Delta S^0_{vap}$

At vaporization point, the liquid and the gaseous phase is in the equilibrium.

Thus,

$\Delta G^0=0$

So,

$Delta H^0_{vap}=T\Delta S^0_{vap}$

Given that:

$Delta H^0_{vap}=55.5\ kJ/mol$

Also, 1 kJ = 10³ J

So,

$Delta H^0_{vap}=55500\ J/mol$

$\Delta S^0_{vap}=148\ J/K.mol$

So, temperature is :

$T=\frac{Delta H^0_{vap}}{\Delta S^0_{vap}}$

$T=\frac{55500}{148}$

3 0

3 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

5 0

3 years ago