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2 years ago

Starch is a type of: Protein Fat Nucleic acid Sugar

Chemistry

1 answer:

Svetllana [295]2 years ago

8 0

Answer:

it's not a protien at all

Explanation:

Starch is not a protein. Starch is a type of carbohydrate. It is a polysaccharide, it has lots of sugars though

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Which of the following samples will have the lowest pressure if they are all at the same temperature and in identical containers

snow_lady [41]

Answer:

c. 15 g Kr

Explanation:

The amount of a gas (Moles) is directely proportional to its pressure. That means the higher amount of moles, the highest pressure and vice versa.

Using molar mass of the compounds (Ne=20.2g/mol, Ar = 39.9g/mol, Kr = 83.8g/mol, CO₂ = 44 g/mol and F₂ = 38.0g/mol), moles of 15.0g of each gas are:

Ne = 15g ₓ (1mol / 20.2g) = 0.74 moles of Ne

Ar = 15g ₓ (1mol / 39.9g) = 0.38 moles of Ar

Kr = 15g ₓ (1mol / 83.8g) = 0.18 moles of Kr

CO₂ = 15g ₓ (1mol / 44g) = 0.34 moles of CO₂

F₂ = 15g ₓ (1mol / 38g) = 0.39 moles of F₂

<h3>As 15g of Kr contains the less quantity of moles, this sample will con have the lowest pressure</h3>

8 0

3 years ago

33. Why does Fluorine exert a stronger pull than Chlorine on the valence electrons of another

mariarad [96]

Answer:

Fluorine is much more reactive than chlorine (despite the lower electron affinity) because the energy released in other steps in its reactions more than makes up for the lower amount of energy released as electron affinity.

Explanation:

7 0

2 years ago

The mass ratio of hydrogen to oxygen in water (formula H20) is 0.126 g hydrogen for every 1 g of oxygen. In another compound of

poizon [28]

Answer: H2O2

Explanation:

5 0

2 years ago

A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0

3 years ago

DUPLICATE. The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to

Law Incorporation [45]

At the half equivalence point [HA] = [A-] and pH = pKa

if Ka is 5.2e-5 then pKa = pH = 4.28

5 0

3 years ago