Chemistry, 14.02.2021 14:00 82coogs

Round off the following measurements to two significant figures. Write the answers in scientific notation. a. 6310 m b. 125.00 Kg c. 6906 km d. 0.0828 m need steps and formela

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Answer from: jdisalle2808

Explanation:

a. 6310 m

Two significant numbers means it must have only 2 digits. The first two digits are 6 and 3. The next number 1 is not up to 5 hence it would be round off to zero.

Two s.f = 6300

Converting it to scientific notation means it must be expressed by a number between 1 and 10, multiplied by a power of 10.

6.3 x 10^3

b. 125.00 Kg

Two significant numbers means it must have only 2 digits. The first two digits are 1 and 2. The next number 5 is up to 5 hence it would be round up to one.

Two s.f = 130.00

Converting it to scientific notation means it must be expressed by a number between 1 and 10, multiplied by a power of 10.

Scientific notation = 1.3 x 10^2

c. 6906 km

Two significant numbers means it must have only 2 digits. The first two digits are 6 and 9. The next number 0 is not up to 5 hence it would be round up to zero.

Two s.f = 6900

Converting it to scientific notation means it must be expressed by a number between 1 and 10, multiplied by a power of 10.

Scientific notation = 6.9 x 10^3

d. 0.0828 m

Two significant numbers means it must have only 2 digits. The first two digits(excluding zero) are 8 and 2 . The next number 8 is up to 5 hence it would be round up to one.

Two s.f = 0.083

Converting it to scientific notation means it must be expressed by a number between 1 and 10, multiplied by a power of 10.

Scientific notation = 8.3 x 10^-2

Answer from: Quest

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

Answer from: Quest

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