How do you make the chemical equations LiOH+H2(SO4) and H2O+Li2(SO4) balanced?

1. In this activity, you will be looking for a relationship between the mass of the cart and the acceleration of the cart.

posledela

Answer:

Mass of cart

Explanation:

7 0

3 years ago

4. The atmosphere is composed of about 78% nitrogen, 21% oxygen, and 1% argon. Typical atmospheric pressure in Boulder, Colorado

photoshop1234 [79]

ANS

Step 1 :Explanation of required formula.

According to Dalton's Law of Partial Pressures, the partial pressure of a component of a gaseous mixture depends on the mole ratio of said component and the total pressure of the gaseous mixture

i.e Pi=Xi × Ptotal,

Here we don't know exactly how many moles of the mixture we have,

but we know that 78.0% of all the molecules present in the mixture are nitrogen molecules, 21.0% are oxygen molecules, and 1% are molecules of Ar gas.

As we know, a mole is simply a very large collection of molecules. In order to have one mole of a substance, we need to have 6.022 × 1023 molecules of that substance.

This means that the actual number of moles is not important here, because the ratio that exists between the number of molecules is equivalent to the ratio that exists between the number of moles.

Hence,

Step 2 : Calculate of mole fraction of the mixture.

mole fraction of nitrogen = 78 /100 = 0.78

mole fraction of O2 =

21 /100 = 0.21

mole fraction of Argon =

1 /100 = 0.01.

Step 3 : Calculate the pressure contributed by each of the mixture.

The pressure contributed by N2 = mole fraction of N2 × Total pressure = 0.78 × 0.83 atm = 0.6474 atm

The pressure contributed by O2 = 0.21 × 0.83 atm = 0.1743 atm

The pressure contributed by N2 = 0.01 × 0.83 atm = 0.0083 atm.

Thank You !!!!!!

6 0

3 years ago

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu

anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = $\frac{3Q}{4\pi R^{3} }$

So, $Q_{1} = \frac{Qr^{3} }{R^{3} }$

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0

3 years ago

Могут ли быть сообщающиеся сосуды неодинаковы пр своей форме

Rudik [331]

Answer:

no

Explanation:

they cannot because they contain the same amount of liquid

8 0

3 years ago

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What quantities are related by Ohm's law? Check all that apply. voltage conductivity current resistance insulation

Free_Kalibri [48]

Answer: Current, resistance and voltage are the quantities which are related by Ohm's law.

Explanation:

A law which states that electric current is directly proportional to voltage and inversely proportional to resistance is called Ohm's law.

Mathematically, it is represented as follows.

$I = \frac{V}{R}$

where,

I = current

V = voltage

R = resistance

This means that the quantities related by Ohm's law include current, voltage and resistance.

Thus, we can conclude that current, resistance and voltage are the quantities which are related by Ohm's law.

7 0

3 years ago

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