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2 years ago

Which trait would most likely lead a scientist to investigate a blood-thinning agent in the saliva of vampire bats?

Chemistry

2 answers:

almond37 [142]2 years ago

4 0

Curiosity because the scientist is curious about the blood thinning agent

Nady [450]2 years ago

3 0

Answer:

Explanation:

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The nucleus of an atom is dense and positively charged. What was observed when positively charged particles were radiated onto a

ankoles [38]

According to my knowledge, I feel the answer is -
Particles that struck the center of the atom were repelled.

Hope this helps!

8 0

2 years ago

Read 2 more answers

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

an unknown metal of mass 512 g at a temperature of 15C is dropped into 325 g of water held in a 100 g aluminum container at the

ziro4ka [17]

Answer:

The specific heat capacity of the metal is 0.843J/g°C

Explanation:

Hello,

To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.

Heat gained by the metal = heat loss by water + calorimeter

Data,

Mass of metal (M1) = 512g

Mass of water (M2) = 325g

Initial temperature of the metal (T1) = 15°C

Initial temperature of water (T2) = 98°C

Final temperature of the mixture (T3) = 78°C

Specific heat capacity of metal (C1) = ?

Specific heat capacity of water (C2) = 4.184J/g°C

Heat loss = heat gain

M2C2(T2 - T3) = M1C1(T3 - T1)

325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)

1359.8 × 20 = 512C1 × 63

27196 = 32256C1

C1 = 27196 / 32256

C1 = 0.843J/g°C

The specific heat capacity of the metal is 0.843J/g°C

8 0

2 years ago

When a known quantity of compound, at a known concentration, is added to a known volume of another compound to determine the con

Vladimir [108]

Answer:

A titration

Explanation:

A common example of a titration is when we have an acid of unknown concentration, so we add a known volume of a base of known concentration. This process lets us determine the concentration of the acid.

By definition, a titration is a quantitative analysis, as we determine how much of an analyte is there in a sample. However, there are quantitative analyzes which are not titrations. This is why the most appropiate answer is a titration.

5 0

2 years ago

Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod

son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0

2 years ago