Read the passage below and choose the best answer for the main idea of the passage:

A tactic used by investigators to identify deceit is to ask the suspect to recount their alibi backwards. True False

MA_775_DIABLO [31]

Answer:

It is True.

Explanation:

Investigators use this tactic to see if the suspect is recalling by memory and if they will change their story after being asked to recall backwards.

8 0

3 years ago

Read 2 more answers

Can you plz help me with this question? thanks! question 6

spayn [35]

Sorry can't see clearly

8 0

3 years ago

What is one way to make a positive impact on air quality to reduce air pollution

geniusboy [140]

Answer:

using public transportation

Explanation:

With the use of public transport, there will be a reduction in the number of cars on the streets, this reduction will cause a reduction in the gases released by cars into the atmosphere, which will result in a reduction in air pollution.

5 0

3 years ago

Experiments with rats show that if rats are dosed with 3.0 mg/kg of cocaine (that is, 3.0 mg cocaine per kg of animal mass), the

Aliun [14]

Answer:

The molecules of dopamine would be produced in a rat after 60 seconds of a 3.0 mg/kg dose of cocaine is $2.258\times 10^{12} molecules$ .

Explanation:

Concentration of dopamine = 0.75 μM = $7.5\times 10^{-7} M$

1 μM = $10^{-6} M$

Volume of the brain in which dopamine is produced = $V=5.00 mm^3$

$1 mm^3=10^{-6} L$

$V = 5.00\times 10^{-6} L$

Moles of dopamine = n

$Concentration=\frac{moles}{Volume}$

$7.5\times 10^{-7} M=\frac{n}{5.00\times 10^{-6} L}$

$n=7.5\times 10^{-7} M\times 5.00\times 10^{-6} L=3.75\times 10^{-12} mol$

$1 mol = 6.022\times 10^{23} molecules$

So, molecules in $3.75\times 10^{-12} mol$ of dopamine :

$3.75\times 10^{-12}\times 6.022\times 10^{23}=2.258\times 10^{12} molecules$

4 0

3 years ago

Find the ph of a buffer that consists of 0.18 m ch3nh2 and 0.73 m ch3nh3cl (pkb of ch3nh2 = 3.35)?

ozzi

Hello!

First, we need to determine the pKa of the base. It can be found applying the following equation:

$pKa=14-pKb=14-3,35=10,65$

Now, we can apply the Henderson-Hasselbach's equation in the following way:

$pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04$

So, the pH of this buffer solution is 10,04

Have a nice day!

8 0

3 years ago