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gayaneshka [121]
1 year ago
13

Express each quantity in regular notation along with its appropriate unit. a. 3.60 × 1 0 5 s 3.60×10 5 s b. 5.4 × 1 0 − 5 g / c

m 3 5.4×10 −5 g/cm 3 c. 5.060 × 1 0 3 k m 5.060×10 3 km d. 8.9 × 1 0 10 H z 8.9×
Chemistry
1 answer:
Gala2k [10]1 year ago
6 0

The numbers in normal notation are as follows:

a. 360000 s

b. 0.000054  g/cm³

c. 5060 km

d. 8900000000 Hz

<h3>What is standard form?</h3>

Standard form is a notation in which numbers are written in such a way that the number is between 1 and 10  multiplied by powers of ten.

Standard form is also known as scientific notation.

The given in numbers are in standard form:

a. 3.60 × 10⁵ s

b. 5.4 × 10⁻⁵ g/cm³

c. 5.060 × 10³ km

d. 8.9 × 10¹⁰ Hz

The numbers in normal notation are as follows:

a. 3.60 × 10⁵ s = 360000 s

b. 5.4 × 10⁻⁵ g/cm³ = 0.000054  g/cm³

c. 5.060 × 10³ km = 5060 km

d. 8.9 × 10¹⁰ Hz = 8900000000 Hz

In conclusion, standard form is also know as scientific notation.

Learn more about standard form at: brainly.com/question/19682432

#SPJ1

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The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the
IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

4 0
2 years ago
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
2 years ago
What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?
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Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

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