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jeka57 [31]
4 years ago
5

Why did Dalton think it was important to use his system of symbols for the chemical elements?

Chemistry
2 answers:
ZanzabumX [31]4 years ago
7 0

Answer:

the answer is a

Explanation:

i did the assignment and the correct answer is a. (i put b bcos other ppl here said its b but i got it wrong)

77julia77 [94]4 years ago
4 0
The answer is : A good luck :)
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Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56 g of NaCl in 164.2 g of H2O. Express your answer to
denis-greek [22]

Answer:

"23.896%" is the right answer.

Explanation:

The given values are:

Mass of NaCl,

= 51.56 g

Mass of H₂O,

= 165.6 g

As we know,

⇒  Mass of solution = Mass \ of \ (NaCl+H_2O)

                                 = 51.56+164.2

                                 = 215.76 \ g

hence,

⇒ Mass \ percent =\frac{Mass \ of \ NaCl}{Mass \ of \ solution}\times 100

                           =\frac{51.56}{215.76}\times 100

                           =23.896 \ percent

4 0
3 years ago
If 175 g of phosphoric acid reacts with 150.0 g of sodium hydroxide, what is the limiting reactant? How many grams of sodium pho
Evgen [1.6K]

Answer:

NaOH is the limiting reactant.

204.9 g of sodium phosphate are formed.

51.94 g of excess reactant will remain.

Explanation:

The reaction that takes place is:

  • H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

First we <u>convert the mass of both reactants to moles</u>, using their <em>respective molar masses</em>:

  • H₃PO₄ ⇒ 175 g ÷ 98 g/mol = 1.78 mol
  • NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol

1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.

--

We <u>calculate the produced moles of Na₃PO₄</u> using the <em>limiting reactant</em>:

  • 3.75 mol NaOH * \frac{1molNa_3PO_4}{3molNaOH} = 1.25 mol Na₃PO₄

Then we <u>convert moles into grams</u>:

  • 1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g

--

We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:

  • 3.75 mol NaOH * \frac{1molH_3PO_4}{3molNaOH} = 1.25 mol H₃PO₄

We substract that amount from the original amount:

  • 1.78 - 1.25 = 0.53 mol H₃PO₄

Finally we <u>convert those remaining moles to grams</u>:

  • 0.53 mol H₃PO₄ * 98 g/mol = 51.94 g
3 0
3 years ago
Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.
e-lub [12.9K]
In order to find the answer, use an ICE chart:

Ca(IO3)2...Ca2+......IO3- 
<span>some.......0..........0 </span>
<span>less.......+x......+2x </span>
<span>less........x.........2x 
</span>
<span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³
</span>
K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
<span>x = molar solubility = 5.6 x 10</span>⁻³ M

The answer is 5.6 x 10 ^ 3 M. (molar solubility)
5 0
3 years ago
Read 2 more answers
Really need help with this! Chemistry
Westkost [7]

Answer:

a) 0,5

Explanation:

If x=6 and y=2, then (2x-4y)/(x+y)=(2*6-4*2)/(6+2)=(12-8)/8=4/8= 0,5

5 0
2 years ago
Chloroform has a density of 1.5 g/ml. What the mass of 10.0 ml of Chloroform?
8090 [49]

Answer:

The answer is 15 g

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

density of Chloroform = 1.5 g/ml

volume = 10 mL

We have

mass = 1.5 × 10

We have the final answer as

<h3>15 g</h3>

Hope this helps you

8 0
3 years ago
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