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4 years ago

The neutrons are also a heavy particle located in the nucleus. these particles contain what particles?

Chemistry

1 answer:

erastova [34]4 years ago

6 0

I think you wanted to ask for charge.

Neutrons does not contain any charge

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A gas occupies 1200 litres at 2 atm pressure. To what pressure must it be compressed to occupy 60 litres at the same temperature

MArishka [77]

P2 = 40 atm

Explanation:

Given:

P1 = 2 atm

V1 = 1200 L

V2 = 60 L

P2 = ?

Using Boyle's law and solving for P2,

P1V1 = P2V2

P2 = (V1/V2)P1

= (1200 L/60 L)(2 atm)

= 40 atm

8 0

3 years ago

How many grams of sodium chromate, Na2CrO4, are needed to react completely with

JulsSmile [24]

Answer:

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

Explanation:

Step 1: data given

Mass of silver nitrate AgNO3 = 56.7 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass of Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 56.7 grams / 169.87 g/mol

Moles AgNO3 = 0.334 moles

Step 4: Calculate moles Na2CrO4 moles needed

For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

Step 5: Calculate mass Na2CrO4

Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

Mass Na2CrO4 = 27.0 grams

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

8 0

3 years ago

The temperature of a star determines its (science)

Anna007 [38]

The answer would be color. The temperature affects the brightness and the shade of the star.

4 0

3 years ago

What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when v

ad-work [718]

Explanation:

At each reflecting surface (benzene and glass) there will be 180 degree phase change.

Now, for constructive interference the optical path in benzene is $\lambda$ .

Formula to calculate thickness of a benzene thin film is as follows.

Optical path length through benzene ( $\lambda$ ) = $2 \times n \times d$

Hence, substituting the given values into the above formula as follows.

Optical path length through benzene = $2 \times n \times d$

d = $\frac{\text{Optical path length through benzene}}{2 \times n}$

= $\frac{\lambda}{2 \times n}$

= $\frac{615 \times 10^{-9}}{2 \times 1.501}$ (as 1 nm = $10^{-9}m$

= 204.9 m

Thus, we can conclude that minimum thickness of benzene is 204.9 m.

4 0

3 years ago

What is the mass of a proton

Anni [7]

1.6726219 × 10-27 kilograms
(The -27 is a exponent)

6 0

3 years ago