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ludmilkaskok [199]
2 years ago
13

Can someone please answer me question 4a only

Mathematics
1 answer:
bixtya [17]2 years ago
4 0

Answer:

b

Step-by-step explanation:

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A 100 centimeter stick broke into 3 pieces one piece was 7 centimeters long and another was 34 centimeters long how long was the
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The third piece is 59 cm long. 

Add the first two pieces up. 7+34= 41. Since the whole stick is 100 cm long, subtract 41 from 100 to find the length of the third piece. 100-41=59. 
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A bakery sold a total of 400 cupcakes in a day, and 144 of them were vanilla flavored. What percentage of cupcakes sold that day
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i think it should be 36%

Step-by-step explanation:

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7 0
3 years ago
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

4 0
3 years ago
The nature center has a fish tank
ANEK [815]

Answer:

288 small fish.

Step-by-step explanation:

The first thing is to calculate the volume of the prism, which would be the area of the base (length by width) and that by the height, that is:

a = l * w * h

l = 6

w = 4

h = 4

Replacing

a = 6 * 4 * 4

a = 96 cubic feet

they tell us that per cubic foot there is room for 3 small fish, therefore:

96 * 3 = 288

Which means that there are 288 small fish.

4 0
3 years ago
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