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3 years ago

What is the difference between diatomic and poly-atomic molecules? Site some examples.

1 answer:

8 0

Their difference lies in the prefix used before each root word, atom. The diatomic molecules are composed only of 2 atoms such as that of the diatomic elements: H2, O2, N2, Cl2, etc. On the other hand, polyatomic molecules are made up of three or mole atoms such as NH4+, H2SO4, etc.

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A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp

Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

P1 = 101 kPa , V1 = 2 m³ , T1 = 300.15 K , P2 = 40 kPa , T2 = 283.15 K

( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

101 × 2 / 300.15 = 40 × V2 / 283.15

V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

V2 ≈ 4.7 m³

7 0

2 years ago

PLEASE HELP ASAP

DaniilM [7]

Your answers:

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Q2. Their food proportions, do they eat every meal, do they eat their vegetables, fruits grains, protein, etc. Check for healthy and unhealthy habits of eating and exercise. Additional notes on medical problems relating to food concerns you may have to look further into based on her answers.

Q3. The info isn't provided, but what I do know is that detection requires awareness of risk factors for, and symptoms and signs of, anorexia nervosa (e.g., participation in activities valuing thinness, family history of an eating disorder, amenorrhea, lanugo hair) and bulimia nervosa (e.g., unsuccessful attempts at weight loss, history of childhood sexual abuse, family history of depression, erosion of tooth enamel from vomiting, parotid gland swelling, and gastroesophageal reflux). Providers must also remain alert for disordered eating in female athletes (the female athlete triad) and disordered eating in diabetics. Treatment requires a multidisciplinary team including a primary care practitioner, nutritionist, and mental health professional. The role of the primary care practitioner is to help determine the need for hospitalization and to manage medical complications (e.g., arrhythmias, refeeding syndrome, osteoporosis, and electrolyte abnormalities such as hypokalemia).

I'm sorry I cannot answer all your questions because there's a typing limit. :)

Thank you for your patience, and I hope I helped!

Would appreciate Brainliest ;)

4 0

3 years ago

A baseball is located at the surface of the earth. Which statements about it are correct? (There may be more than one correct ch

lakkis [162]

i think that the answer is a)

3 0

3 years ago

Brainliest if right

mixas84 [53]

They traveling at -0.37/ms^

3 0

2 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago