A melting point of over 700 C and a density of less than 2 g/cm3 can be observed for many group 2 elements. In this group, the density increases on moving down the group, whereas the melting point increases upto calcium and then starts decreasing.
Calcium, symbol Ca is the element with melting point around 840 C and density of 1.55 g/cm3 which is closest to the specified data range .
Answer:
Fossils, Glaciers, Coastlines.
Explanation:
Evidence of plate tectonic theory are fossils, fossils of the same species were found oceans apart. also glaciers evidence of glaciers were found in places with warm climates. coastlines coastlines look like they fit in a puzzle.
Answer: -
12.59
Explanation: -
Strength of NaOH = 0.0179 M
Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L
Number of moles = 0.0179 M x 0.058 L
= 1.04 x 10⁻³ mol
Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol
Strength of Ba(OH)₂ = 0.0294 M
Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L
Number of moles = 0.0294 M x 0.060 L
= 1.76 x 10⁻³ mol
Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol
Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol
= 4.56 x 10⁻³ mol
Total volume of the mixture = 58.0 + 60.0
= 118.0 mL
118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]
1000 mL of the solution has 
= 0.0386 mol
Using the relation
pOH = - log [OH-]
= - log 0.0386
= 1.41
Using the relation
pH + pOH = 14
pH = 14 - 1.41
= 12.59
Answer:
removing a top predator may cause its prey to become more abundant, as fewer individuals are eaten. But with more prey around, the organisms that it eats may become scarcer. Seemingly simple changes can have complex effects, with direct and indirect interactions rippling throughout entire ecosystems.
Explanation:
Answer:
Number of molecules = 1.8267×10^20
Explanation:
From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;
PV = nRT
where
P =pressure
V =volume
n = the number of moles
R is the gas constant equal to 0.0821 L·atm/mol·K
T is the absolute temperature
Given:
P = 6.75 atm;
T = 290.0 k,
; V = 1.07 cm³ = 0.001 L
( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)
n = 3.0335167*10^-4 moles
But there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 1.8267×10^20