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3 years ago

What occurred when rutherford exposed nitrogen gas to alpha particles?

2 answers:

5 0

When Rutherford exposed nitrogen gas to alpha particles, some of the alpha particles were absorbed by the nitrogen nuclei.

A equation for the alpha particle bombardment is as follows:

¹⁴₇N+⁴₂He---->¹⁷₈O+¹₁H

So, as it can be seen from the equation that some of the alpha particles are absorbed by nitrogen nuclei.

adoni [48]3 years ago

5 0

The conversion of one element to another had occurred.

Please mark me brainiest.

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Round off each number to the indicated number significant figures (sf):

qaws [65]

Answer :

(a) The given number in 2 significant figures will be, $3.6\times 10^3$

(b) The given number in 4 significant figures will be, $3.583\times 10^1$

(c) The given number in 3 significant figures will be, $2.24\times 10^1$

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

(a) The given number 00003554 converted into 2 significant figures.

In 00003554, there are 4 significant figures. Now we have to convert it into 2 significant figures.

The given number in 2 significant figures will be, $3.6\times 10^3$

(b) The given number 35.8348 converted into 4 significant figures.

In 35.8348, there are 6 significant figures. Now we have to convert it into 4 significant figures.

The given number in 4 significant figures will be, $3.583\times 10^1$

(c) The given number 22.4555 converted into 3 significant figures.

In 22.4555, there are 6 significant figures. Now we have to convert it into 3 significant figures.

The given number in 3 significant figures will be, $2.24\times 10^1$

7 0

3 years ago

Question 2 (2 points) Question 2 Unsaved Why does a blue shirt look blue? Choose the BEST explanation. Question 2 options: All w

Mazyrski [523]

1 . Each color has a different wavelength allowing the eye to see it.2 . The shirt reflects the blue wavelengths.
3 . Charcoal absorbs all wavelengths of light that fall on it4 . Red
5 . Purple 6 . Yellow7 . It contains all of the wavelengths of the visible light spectrum.

5 0

3 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

2 years ago

In an experiment, marble chips reacted with dilute hydrochloric acid and carbon dioxide gas was given

laiz [17]

Answer:

Explanation:

Small marble chips reacts faster

7 0

3 years ago

How many anions are in 0.500 g of MgBr2

Oksana_A [137]

Given:

Mass of MgBr2 = 0.500 g

To determine:

Number of anions in 0.500 g MgBr2

Explanation:

Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol

Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles

Based on stoichiometry-

1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions

Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-

Now,

1 mole of Br- contains 6.023 * 10²³ anions

0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions

Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2

4 0

3 years ago