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Aloiza [94]
3 years ago
6

How many fiftieths would three tenths equal

Mathematics
2 answers:
sashaice [31]3 years ago
7 0
Convert 3/10 to a fraction over 50.

50 / 10 = 5

Multiply both the numerator and denominator by 5:

5 * 3 = 15
5 * 10 = 50

15/50

So 15 fiftieths would equal three tenths.
san4es73 [151]3 years ago
3 0
Use ratio and proportion

x/50=3/10
cross multiply
10x=150
     x=15
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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
Which equation represents the line that passes through the points (-5,2) and (9,0)
V125BC [204]

Step-by-step explanation:

<em>Given </em><em>points </em>

<em>(</em><em> </em><em>-</em><em>5</em><em> </em><em>,</em><em> </em><em>2</em><em>)</em><em> </em><em>and </em><em>(</em><em>9</em><em> </em><em>,</em><em> </em><em>0</em><em>)</em>

<em>First </em><em>calculating </em><em>slope(</em><em>m</em><em>)</em>

<em>=</em><em> </em><em>(</em><em> </em><em>y2 </em><em>-</em><em> </em><em>y1</em><em>) </em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>x2 </em><em>-</em><em> </em><em>x1</em><em>) </em>

<em>=</em><em> </em><em>(</em><em> </em><em>0</em><em>-</em><em>2</em><em>)</em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>9</em><em>+</em><em>5</em><em>)</em>

<em>=</em><em> </em><em>-</em><em>2</em><em> </em><em>/</em><em> </em><em>1</em><em>4</em>

<em>=</em><em> </em><em>-</em><em>1</em><em>/</em><em>7</em>

<em>Now </em><em>The </em><em>equation </em><em>of </em><em>line </em><em>is </em><em>given </em><em>by </em>

<em>y </em><em>-</em><em> </em><em>y1 </em><em>=</em><em> </em><em>m </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>x1) </em>

<em>y </em><em>-</em><em>2</em><em> </em><em>=</em><em> </em><em>-</em><em>1</em><em>/</em><em>7</em><em> </em><em>(</em><em>x </em><em>+</em><em> </em><em>5</em><em>)</em>

<em>7</em><em>(</em><em> </em><em>y </em><em>-</em><em> </em><em>2</em><em>)</em><em> </em><em>=</em><em> </em><em>-</em><em>1</em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>5</em><em>)</em>

<em>7y </em><em>-</em><em> </em><em>1</em><em>4</em><em> </em><em>=</em><em> </em><em>-</em><em>x </em><em>-</em><em>5</em>

<em>x </em><em>+</em><em> </em><em>7y </em><em>-</em><em>1</em><em>4</em><em> </em><em>+</em><em>5</em><em> </em><em>=</em><em> </em><em>0</em>

<em>x </em><em>+</em><em> </em><em>7y </em><em>-</em><em> </em><em>9</em><em> </em><em>=</em><em> </em><em>0</em><em> </em>

<em>Which </em><em>is </em><em>the </em><em>required </em><em>equation </em>

7 0
3 years ago
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