Is it technically correct to call ionic compounds molecules?

If an equilibrium mixture of the three gases at 600K contains 2.92*10^-2 M COCH(g) and 1.76*10^2 M CO, what is the equilibrium

Vlad [161]

Answer:

C

Explanation:

7 0

3 years ago

For each pair, predict which molecule has the greater molar entropy under the same conditions (assume gaseous species).c o 2, ca

larisa [96]

I used to know this. I don't really remember but is used to know it. Im in 6th grade now

8 0

4 years ago

If one or more nucleotide pairs are deleted from a dna strand, this is known as a:_________

Leya [2.2K]

If one or more nucleotide pairs are deleted from a DNA strand, this is known as a frameshift mutation

<h3>Define Frameshift Mutation</h3>

Insertions or deletions in the genome that are not multiples of three nucleotides are referred to as frameshift mutations. They are a particular class of insertion-deletion (indel) alterations that are present in polypeptides' coding sequences. Here, there are no multiples of three in the number of nucleotides that are added to or subtracted from the coding sequence. They may result from really basic alterations like the insertion or deletion of a single nucleotide.

<h3>Frameshift mutations' effects</h3>

One of the most harmful modifications to a protein's coding sequence is a frameshift mutation. They are quite prone to produce non-functional proteins that frequently interfere with a cell's metabolic processes and result in significant alterations to polypeptide length and chemical makeup. Frameshift mutations can cause the mRNA to stop translating too soon and create an extended polypeptide.

Learn more about Frameshift mutations here:-

brainly.com/question/12732356

#SPJ4

4 0

1 year ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

3 years ago

Antarctica, almost completely covered in ice, has an area

Marysya12 [62]

<u>Answer:</u> The mass of ice is $2.39\times 10^{22}g$

<u>Explanation:</u>

We are given:

Area of Antarctica = $5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2$ (Conversion factor: $1mi^2=2.59\times 10^{10}cm^2$ )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = $182.88\times 10^3cm$ (Conversion factor: 1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Density of ice = $0.917g/cm^3$

Volume of ice = Area × Height of ice = $142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3$

Putting values in above equation, we get:

$0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g$

Hence, the mass of ice is $2.39\times 10^{22}g$

5 0

3 years ago