All categories

2 years ago

How does our understanding of genetics influence psychological research

1 answer:

8 0

I think that an understanding of genetics and traits and dispositions that are handed down through heredity can allow us to study such traits in their psychological aspects ie people with a very high metabolism can pass this on to their offspring and there may be some correlation between this and having a manic disposition.

You might be interested in

The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri

Nimfa-mama [501]

Answer: The expression for equilibrium constant is $\frac{[NH_3]^2}{[H_2]^3[N_2]}$

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as $k_c$

For a general reaction:

$aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is written as:

$k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Chemical reaction for the formation of ammonia is:

$N_2+3H_3\rightleftharpoons 2NH_3$

$k_c=1.6\times 10^2$

Expression for $k_c$ is:

$k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

$1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

8 0

2 years ago

What changes are observed on heat capacity ratio and output temperatures by increasing the specific heat capacity of both the fl

BlackZzzverrR [31]

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

8 0

3 years ago

What must be true of a disproportionate substance?

Rudik [331]

Answer:

Answer is D.it gains and loses electrons.

Explanation:

I hope it's helpful!

4 0

2 years ago

Read 2 more answers

Consider the diagram below

Artyom0805 [142]

Energy of the reactants would be correct

5 0

1 year ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

2 years ago