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3 years ago

In living things genetic material is stored in? A. Lipids B. Amino Acids C. Nucleic Acids

Chemistry

2 answers:

swat323 years ago

4 0

B- Nucleic acid is the answer

miv72 [106K]3 years ago

3 0

C- Nucleic Acids is the correct answer :) DNA and RNA

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What is produced when calcium reacts with fluorine in a synthesis reaction?

Thepotemich [5.8K]

C is the correct answer (CaF2) (sorry dont have subscript)

Explanation: synthesis reaction forms a compound and calcium reacting with fluorine produces Calcium Fluoride (CaF2) chemical name

4 0

3 years ago

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What element is likely to form an ionic compound with chlorine?

kirill [66]

Sodium (NA)
the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.

5 0

3 years ago

Is this correct?<br> Plz answer ASAP!

Angelina_Jolie [31]

YES IT'S @LL CORRECT

3 0

3 years ago

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The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of

navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

$M=\frac{n}{V}$

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

$2.60g*\frac{1mol}{102.89g} =0.0253mol$

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

$M=\frac{0.0253mol}{0.160L}=0.158M$

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

$V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L$

That in milliliters turns out to be:

$V=0.211L*\frac{1000mL}{1L}=211mL$

Learn more about molarity: brainly.com/question/10053901

6 0

2 years ago