When the nuclide uranium-234 undergoes alpha decay: the name of the product nuclide is . the symbol for the product nuclide is?

a) Thorium

b)

Explanation:

Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.

General representation of an element is given as:  

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

General representation of alpha decay :

Hence the name of the product nuclide is Thorium and the symbol for the product nuclide is

a mercury barometer is a device that uses the physical properties of mercury in a tube to measure barometric pressure. mercury barometers are used for : -

1. measurement of atmospheric pressure  

2. determining the altitude of a place

3. weather forecasting

4. calibration and checking of aneroid barometers

5. application in fluid mechanics

6. measurement of pressure in aircrafts

7. surface weather analysis

8. preparation of barographs

9. usage in physical sciences like physics, astronomy & chemistry

10. preparation of aircraft altimeters

characteristics of mercury barometer : -

1.used to define and measure atmospheric pressure.

2.on the average at sea level the column of mercury rises to a height of about 760 mm.

3.this quantity is equal to 1 atmosphere.

4.it is also known as standard atmospheric pressure.

5.a barometer in which the weight of a column of mercury in a glass tube with a sealed top is balanced against that of the atmosphere pressing on an exposed cistern of mercury at the base of the mercury column, the height of the column varying with atmospheric pressure.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol