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SVETLANKA909090 [29]
3 years ago
15

At room temperature, compounds that are held together by covalent bonds tend to exist in what state?

Chemistry
2 answers:
Sophie [7]3 years ago
6 0

at room temperature substances hawks together by covalent bonds are typically in liquid form

Monica [59]3 years ago
5 0

Answer:

Explanation

since they cannot form 3/d crystals, the molecules flow past each other

l

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If the detector is capturing 3.3×108 photons per second at this wavelength, what is the total energy of the photons detected in
Reil [10]

Answer:

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

Explanation:

The energy carried by electromagnetic radiation is displaced by waves. This energy is not continuous, but is transmitted grouped into small "quanta" of energy called photons. The energy (E) carried by electromagnetic radiation can be measured in Joules (J). Frequency (ν or f) is the number of times a wave oscillates in one second and is measured in cycles / second or hertz (Hz). The frequency is directly proportional to the energy carried by a radiation, according to the equation: E = h.f, (where h is the Planck constant = 6.63 · 10⁻³⁴ J / s).

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. it is expressed in units of length (m). In light and other electromagnetic waves that propagate at the speed of light (c), the frequency would be equal to the speed of light (≈ 3 × 10⁸ m / s) between the wavelength :

f=\frac{speed of light}{wavelength}

So:

E=\frac{h*speed of light}{wavelength}

In this case, the wavelength is 3.35mm=3.35*10⁻³m and the energy per photon is:

E=\frac{6.63*10^{-34}*3*10^{8}}{3.35*10^{-3} }

E=5.93*10⁻²³ \frac{J}{proton}

The detector is capturing  3.3*10⁸ photons per second. So, in 1 hour:

E=5.93*10^{-23} \frac{J}{proton} *3.3*10^{8} \frac{proton}{s} *\frac{60}{1} \frac{s}{minute} *\frac{60}{1} \frac{minute}{hr}

E=7.04*10⁻¹¹ \frac{J}{hr}

The total energy of the photons detected in one hour is 7.04*10⁻¹¹ J

3 0
3 years ago
Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha
Aleks04 [339]

Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

pH = pKa + log \frac{[Base]}{[Acid]}

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

7.5 = 6.3 + log \frac{[base]}{[acid]}

1.2 = log \frac{[base]}{[acid]}

\frac{[Base]}{[Acid]} = 10^{1.2}

\frac{[Base]}{[Acid]} = 15.8

[Base] = 15.8 \times [Acid]

The total of mole fraction of acid and base is 1. Therefore we have,

[Acid] + [Base] = 1

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

[Acid] + 15.8 \times [Acid] = 1

16.8 [Acid] = 1

[Acid] = \frac{1}{16.8}

[Acid] = 0.0595

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

4 0
3 years ago
Read 2 more answers
What is the chemical name of this compound? Na2S:
ahrayia [7]

Answer:

Sodium Sulfuret

7 0
3 years ago
How is motion involved in your lives?​
PolarNik [594]

Answer:

By walking and stuff, duh.

Explanation:

We walk, motion.

We drive, motion.

We eat, motion.

We talk, motion.

8 0
3 years ago
Read 2 more answers
Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 K. Upon combusting t
Masja [62]

Answer:

Molecular formula of hydrocarbon is: C₂H₆

Explanation:

The combusting of the hydrocarbon:

CxHy + O₂ → CO₂ + H₂O

Using gas law to obtain molar mass of the gas mixture (CO₂ + H₂O):

P/RT = n/V

Where:

P is pressure (1,208 atm)

R is gas constant (0,082 atmL/molK)

T is temperature (375 K)

n/V = 0,0393 mol/L

1,1128 g/L ÷ 0,0393 mol/L = 28,32 g/mol

Thus, average molecular weight is:

28,32 g/mol = 44,01 g/mol X + 18,02 g/mol Y

1 = X + Y

Where X is CO₂ molar percentage and Y is H₂O molar percentage.

Solving:

X = 0,397

Y = 0,603

39,7% H₂O

60,3% CO₂

With this proportion you can obtain ratio CO₂:H₂O thus:

60,3/39,7 = 1,52 So, 2 CO₂: 3H₂O

The moles of the hydrocarbon are:

PV/RT = n

P (0,959 atm)

V ( 1/5 of final volume)

T (298K)

n = 7,85x10⁻³ mol

The moles of CO₂ are:

0,0393 mol × 2 mol CO₂/ 5 mol total =0,01572.

Ratio of CO₂:CxHY =

0,01572 : 7,85x10⁻³ 2 CO₂: 1 CxHY

Doing:

1 CxHy + O₂ → 2 CO₂ + 3 H₂O

By mass balance:

1 C₂H₆ + 7/2 O₂  → 2 CO₂ + 3 H₂O

Thus, molecular formula of hydrocarbon is: <em>C₂H₆ </em>

I hope it helps!

7 0
3 years ago
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