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Bogdan [553]
4 years ago
7

I need help ASAP! It's urgent.. PLISSSSS​

Mathematics
1 answer:
natali 33 [55]4 years ago
5 0

Answer:

a) 6 mins

b) 70km/h

c) t= 45

Step-by-step explanation:

a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km

Duration

= 16 -10

= 6 minutes

b) Average speed

= total distance ÷ total time

Total time

= 24min

= (24÷60) hr

= 0.4 h

Average speed

= 28 ÷0.4

= 70 km/h

c) Average speed= total distance/ total time

Average speed

= 80km/h

= (80÷60) km/min

= 1⅓ km/min

1⅓= 28 ÷(t -24)

<em>since</em><em> </em><em>duration</em><em> </em><em>for</em><em> </em><em>return</em><em> </em><em>journey</em><em> </em><em>is</em><em> </em><em>from</em><em> </em><em>t</em><em>=</em><em>2</em><em>4</em><em> </em><em>mins</em><em> </em><em>to</em><em> </em><em>t</em><em> </em><em>mins</em><em>.</em>

\frac{4}{3}(t -24)= 28

\frac{4}{3}t - 32= 28

\frac{4}{3}t= 32 +28

\frac{4}{3}t= 60

t= 6 0\div  \frac{4}{3}

t= 45

*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.

Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)

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PLEASE HELP!!!
Aliun [14]

Answer:

a. 8.75 M NaOH

b. 0.425 M CuCl₂ or 0.43 M CuCl₂

c. 0.067 M CaCO₃ or 0.07 M CaCO₃

Step-by-step explanation:

Molality is computed using the formula:

M = \dfrac{Moles\:of\:solute}{Liters\:of\:solution}

So first thing you need to do is determine how many moles of solute there are and divide it by the solution in liters.

Converting mass to moles, you need to get the mass of each solute per mole. You can use the periodic table to get the atomic mass (which is the grams per mole of each atom) of each of the elements involved. Then add them up and you will have how many grams per mole of each compound.

1. 35.0g of NaOH in 100ml H₂O

Element         number of atoms            atomic mass           TOTAL

Na                              1                   x             22.99g/mol  =    22.99g/mol

O                                1                   x             16.00g/mol   =    16.00g/mol

H                                1                   x                1.01g/mol   =<u>       1.01g/mol</u>

                                                                                                 40.00g/mol

This means that the molecular mass of NaOH is 40.00 g/mol

Then we use this to convert 35.0g of NaOH to moles:

35.0g \:of\:NaOH \times \dfrac{1\:mole\:of\:NaOH}{40.00g\:of\:NaOH} = \dfrac{35.0\:moles\:of\:NaOH}{40.00}=0.875\:moles\:of\:NaOH

Now that you have the number of moles we divide it by the solution in liters. Before we can do that you have to conver 100ml to L.

100ml\times\dfrac{1L}{1000ml} = 0.1 L

Then we divide it:

\dfrac{0.875\:moles\:of\:NaOH}{0.1L of solution} = 8.75M\: NaOH

2. 20.0g CuCl₂ in 350ml H₂O

Element         number of atoms            atomic mass           TOTAL

Cu                              1                   x             63.55g/mol  =    63.55g/mol

Cl                               2                   x             34.45g/mol   =   <u>70.90g/mol</u>

                                                                                                134.45g/mol

20.g\:of\:CuCl_2\times\dfrac{1\:mole\:of\:CuCl_2}{134.45\:g\:of\:CuCl_2}=0.1488\:moles\:of\:CuCl_2

350ml = 0.350L

\dfrac{0.1488\:moles\:of\:CuCl_2}{0.350L\:of\:solution}=0.425M\:CuCl_2

3. 3.35g CaCO₃ in 500ml

Element         number of atoms            atomic mass           TOTAL

Ca                              1                   x             40.08g/mol  =    40.08g/mol

C                                1                   x              12.01g/mol   =    12.01g/mol

O                                3                  x              16.00g/mol   =<u>   48.00g/mol</u>

                                                                                                100.09g/mol

3.35g\:of\:CaCO_3\times\dfrac{1\:mole\:of\:CaCO_3}{100.09\:g\:of\:CaCO_3}=0.0335\:moles\:of\:CaCO_3

500ml = 0.5L

\dfrac{0.0335\:moles\:of\:CaCO_3}{0.5L}=0.067M\:CaCo_3

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