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seraphim [82]
3 years ago
9

Refer to the picture of the satellite orbiting the Earth to help you answer this question.

Chemistry
1 answer:
anastassius [24]3 years ago
6 0

Explanation:

D. satellite d

...............

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The cost of 1kg of potatoes is $8. What is the cost of 15 kg of potatoes? (hurry pls)
Ostrovityanka [42]
It’s 8•15 which is 120 kg of potatoes
6 0
3 years ago
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NH3(aq) + HNO3(aq) → NH4NO3(aq) Calculate the volume of an acid (1.5 M HNO3) needed to neutralize the 1.5 M HNO3.
SCORPION-xisa [38]

Answer:

NH3(aq) + HNO3(aq) → NH4NO3(aq) Calculate the volume of an acid (1.5 M HNO3) needed to neutralize the 1.5 M HNO3.

Explanation:

8 0
3 years ago
Write the appropriate symbol for each of the following isotopes: (a) Z 74, A 186; (b) Z 80, A 201; (c) Z 34, A 76; (d) Z 94, A 2
beks73 [17]

Answer:  a)  _{74}^{186}\textrm{W}

b) _{80}^{201}\textrm{Hg}

c)  _{34}^{76}\textrm{Se}

d) _{94}^{239}\textrm{Pu}

Explanation:

General representation of an element is given as:_Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

In an atom, when neutrons or protons are lost or gains, it directly affects the mass number of an atom.

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

It is characteristic of a particular element.

Atomic number = Number of electrons = Number of proton

a) Z 74, A 186: _{74}^{186}\textrm{W}

b) Z 80, A 201: _{80}^{201}\textrm{Hg}

c) Z 34, A 76: _{34}^{76}\textrm{Se}

d) Z 94, A 239.: _{94}^{239}\textrm{Pu}

6 0
3 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0
3 years ago
What is the unabbreviated electron configuration for Neptunium?
slava [35]

Answer:

Rn 5f4 6d1 7s2

Explanation:

6 0
2 years ago
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