4) During convection, what happens to a fluid? Hot fluids will rise, while cold fluids sink.
Cold fluids will rise, while hot fluids sink.
Both hot and cold fluids will rise.
Both hot and cold fluids will sink.

it's the first one

Explanation:

hot fluids rise out the cold fluids

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

wanahton purified a portion of water with 900 grams of contaminants. each hour, a third of the contaminants was filtered out.

let g(n) be the amount of contaminants (in grams) that remained by the beginning of the nth hour. g is a sequence, what kind of sequence is it, arithmetic or geometric? then, write an explicit formula for the sequence, g(n)=

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