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ioda
3 years ago
11

How long must $542 be invested at a rate of 7% to earn $303.52 in interest

Chemistry
1 answer:
yulyashka [42]3 years ago
7 0

The number of years that must be invested at  a rate of 7 % to earn$ 303.52 in interest is

    8 years

 <em><u>calculation</u></em>

  • <em><u> </u></em><em>by use of  the formula       A = P (1+  rt) </em>
  • <em>  where :  A  is the final amount  =  542 + 303.52  =$ 845.52</em>

<em>                    P  is the principal   money to be invested =  $ 542</em>

<em>                        r= rate= 7/100=0.07</em>

<em>                       t=  time required</em>

<em>=$ 845.52=$ 542( 1+  0.07 t)</em>

  • <em>open the  bracket</em>
  • <em>= $845.53= $542 + $37.94  t</em>

  • <em>like terms  together</em>

    =$  845.53 -$542  = $37.94 t

   =$303.52 =$37.94 t

  • divide both  side by  $37.94

  =  $303.52/ $   37.94  =  $37.94t/$37.94

  t=  8 years

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If an element has an atomic number of 20 and a mass number of 41, what is the :
iVinArrow [24]

Answer:

20 protons, 20 electrons, and 21 neutrons

Explanation:

The atomic number of an atom is the number of protons it has. If the atomic number is 20 then we know the atom has 20 protons.

•The mass number of an atom is the total number of protons and neutrons the atom contains. The mass number is 41 and the number of protons is 20, just subtract 20 from 41 and you will get the number of neutrons: 41 - 20= 21. The atoms has 21 neutrons.

•The number of electrons found in an atom is equal to the number of protons. The atoms has 20 protons which means it has 20 electrons.

So, the answer is:

20 protons, 20 electrons, and 21 neutrons

8 0
3 years ago
Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -&gt; 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

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

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
3 years ago
How long does it take the Earth to move from Point A to Point C?
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50 POINTS for a legitimate response. Please answer the question, chose the correct unit conversions on the right. Please read th
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Suppose we have two rock samples, A and B. If rock A was subject to both physical and chemical weathering while rock B was subje
hammer [34]

Answer:

Rock A because after physical weather and chemical weathering it more likely for Rock A to experience more chemical weathering.

Explanation: Weathering: This is a geological term used to describe the various processes and Activities involved in the breaking down of rocks either through physical,mechanical,chemical etc actions into smaller particles.

ROCK A WILL HAVE MORE CHEMICAL WEATHERING BECAUSE THE PHYSICAL WEATHERING MUST HAVE BROKEN DOWN THE PARTICLES FOR EASY WATER AND OTHER SUBSTANCE NEEDED FOR EASIER CHEMICAL REACTION OR WEATHERING.

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3 years ago
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