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zalisa [80]
3 years ago
7

The clouds of Venus consist mainly of ______. Group of answer choices carbon monoxide droplets of water vapor hydrogen and heliu

m droplets of sulfuric acid gaseous nitrogen
Chemistry
1 answer:
Zanzabum3 years ago
4 0

Answer:

droplets of sulfuric acid

Explanation:

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On a molar basis, which is a more effective bleaching agent- kbro3 or naocl?
Aleksandr-060686 [28]
<span>On a molar a basis, the most effective bleaching agent kbro3 has more oxygen, but due to the bromine, it would make things look yellowish. The bromine would counteracts the bleaching effect. As we all know, bleach is naocl or hypochlorite</span>
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3 years ago
How many molecules are in 4.2 miles of water?​
viktelen [127]

Answer:

2.5284 x 10^24

4 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
Calculate the molarity of the two solutions.
daser333 [38]

1. 0.33 M

2. 0.278 M

<h3>Further explanation</h3>

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

1. 0.350 mol of NaOH in 1.05 L of solution.

n=0.35

V=1.05 L

Molarity :

\tt M=\dfrac{0.35}{1.05}=0.33

2. 14.3 g of NaCl in 879 mL of solution.

mol NaCl(MW=58.5 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{14.3~g}{58.5~g/mol}=0.244

Molarity :

\tt M=\dfrac{0.244}{0.879~L}\\\\M=0.278

4 0
2 years ago
What is boiling point​
frutty [35]

Answer:

Its the temperature at which the molecules of a matter especially a liquid attain higher kinetic energy and the rate of collision becomes increased

Explanation:

8 0
3 years ago
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