An object has a mass of 26.94 grams and a volume of 2.568 cubic centimeters. What material is it likely to be made of?
Substance iron gold silver copper
Density (g/cm) 7.874 19.32 10.49 8.92
a. Silver
c. Iron
b. Gold
d. Copper

a) silver

Explanation:

Given data:

Mass of object = 26.94 g

Volume of object = 2.568 cm³

Which subject is this = ?

Solution:

Formula:

d = m/v

by putting values

d = 26.94 g/ 2.568 cm³

d =  10.49 g/cm³

Densities of objects: g/cm³

iron =  7.874

gold = 19.32

silver =  10.49

copper = 8.92

Thus, given object is silver.

mechanics (greek   is that area of science concerned with the behaviour of physical bodies when subjected to forces or displacements, and the subsequent effects of the bodies on their environment. the scientific discipline has its origins in ancient greece with the writings of aristotle and archimedes[1][2][3] (see history of classical mechanics and timeline of classical mechanics). during the early modern period, scientists such as galileo, kepler, and newton laid the foundation for what is now known as classical mechanics. it is a branch of classical physics that deals with particles that are either at rest or are moving with velocities significantly less than the speed of light. it can also be defined as a branch of science which deals with the motion of and forces on objects. the field is yet less widely understood in terms of quantum theory.

explanation:

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