A sample of neon gas occupies a volume of 752 mL at

Temka [501]

1. Ca(HCO3)2
2.Ca(HCOO)2
3. Ca(OH)2
4.NaOH
5.KCI
6.MgSO4
7.PbO
8.HCl
9.HNO3
10.H2SO4
11.NH3
12.(NH4)3PO4
13.NaOH
:)

5 0

2 years ago

True or false: Driving a motor vehicle after taking substances which alter The central nervous system can have side effects whic

Norma-Jean [14]

True, because the substance basically reacts to the nervous system just like the anesthetic. Which can affect your decision making, and you might experience drowsiness, slow reaction time which is very risky for the driver and his/her surroundings.

Hope this helps :)

4 0

3 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago

PLEASE!!! I NEED HELP ASAP!!

Lana71 [14]

W=m₁/m₀=2^(-t/T)

t=4.6·10⁹ years
T=5·10¹⁰ years

w=2^(-4.6·10⁹/5·10¹⁰)

w=0.9382
w=93.82%

7 0

2 years ago

Read 2 more answers

If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of

navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

$M=\frac{n}{V}$

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

$2.60g*\frac{1mol}{102.89g} =0.0253mol$

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

$M=\frac{0.0253mol}{0.160L}=0.158M$

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

$V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L$

That in milliliters turns out to be:

$V=0.211L*\frac{1000mL}{1L}=211mL$

Learn more about molarity: brainly.com/question/10053901

6 0

1 year ago