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Ivan
3 years ago
8

The viscosities of several liquids are being compared. All the liquids are poured down a slope with equal path lengths. The liqu

id with the highest viscosity will _____.reach the bottom firstreach the bottom last
Chemistry
1 answer:
boyakko [2]3 years ago
8 0

Explanation:

The property of a substance to resist the flow of motion is known as viscosity. And, more is the density of a substance more will be its viscosity.  

Whereas, lesser is the density of a substance then it is easy for the substance to move.

This means that more is the viscosity of a substance least will be its flow and when a substance has lesser viscosity then it will readily flow from one point to another.

Thus, we can conclude that the viscosities of several liquids are being compared. All the liquids are poured down a slope with equal path lengths. The liquid with the highest viscosity will reach the bottom last.

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Which of the following is the smaller atom: magnesium or chlorine?
Aleks04 [339]

Answer:

Chlorine atoms are smaller

Explanation:

Magnesium have more electrons than chlorine.

3 0
3 years ago
Read 2 more answers
(20 Pt.s) Guys! You have to explain:
Nadya [2.5K]

Weakly basic drugs behaves different from acidic drugs which is discussed below.

<h3><u>Explanation</u>:</h3>

Weakly basic drugs are those drugs which have an amine group associated with them. They are able to gain a proton to be come positively charged.

So drugs like quinine, ephedrine and aminopyrine which are basic got completely ionised in stomach.

The stomach can absorb those compounds which are lipid soluble. The acidic drugs like alcohols, Salicylic acid, aspirin, thiopental, secobarbital and antipyrine etc which are acidic gets absorbed by means of <em>diffusion</em> through the membrane.

But the basic drugs have charges on them which makes them lipophobic. So they cannot get absorbed through stomach. However weakly basic drugs sometimes get absorbed depending on their ionisation extent.

The rest goes to small intestine which has basic environment and there they gets absorbed via diffusion or facilitated diffusion.

3 0
3 years ago
Find the number of atom in 1.50g of c
Aleonysh [2.5K]
Atomic mass Carbon (C ) = 12.01 a.m.u

12.01 g ---------- 6.02x10²³ atoms
1.50 g ----------- ??

1.50 x ( 6.02x10²³ ) / 12.01 =

7.51x10²² atoms of C

hope this helps!

6 0
3 years ago
CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
The reaction below will occur in a gaseous system at STP:
WARRIOR [948]

Answer:

V = 12.5 L

Explanation:

Given data:

Volume of NO = 15.0 L

Temperature and pressure = standard

Volume of nitrogen gas produced = ?

Solution:

Chemical equation:

6NO + 4NH₃    →     5N₂ + 6 H₂O

Number of moles of NO:

PV = nRT

n = PV/RT

n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K

n = 15.0 atm.L / 22.43 atm.L /mol

n = 0.67 mol

now we will compare the moles of No and nitrogen gas.

              NO           :         N₂

               6             :          5

              0.67         :         5/6×0.67 = 0.56

Volume of nitrogen gas:

 PV = nRT

1 atm × V = 0.56 mol ×  0.0821 atm.L /mol.K × 273.15 K

V = 12.5 atm.L / 1 atm

V = 12.5 L

4 0
3 years ago
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