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balandron [24]
2 years ago
12

11 What is the ionic charge for an ion with 12 p* and 10 e? A 0 B +2 C -2 D +1

Chemistry
1 answer:
aliina [53]2 years ago
3 0

Answer:

+2

Explanation:

10 e   will 'balance '   10 p    then there is two + charges leftover

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The balanced equation is 2Cr + 3Cl2 = 2CrCl3
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Why was Becquerel’s experiment a first step in discovering radioactivity?
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This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
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The expected value for a chemical equation is 47g of water, after an experiment you find that you have 2.58 moles of water. What
strojnjashka [21]
<h3><u>Answer</u>;</h3>

Actual yield = 46.44 g

<h3><u>Explanation;</u></h3>

1 mole of water = 18 g/mol

Therefore;

The experimental yield = 2.58 moles

equivalent to ; 2.58 × 18 = 46.44 g

The theoretical value is 47 g

Percentage yield = 46.44/47 × 100%

                             = 98.8%

The questions asks for actual yield = 46.44 g

6 0
3 years ago
"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum
UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = 25^{o} C

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = C \times 0.082 \times 298.15 K&#10;

               C = 0.0285

This also means that,

  \frac{\text{moles}}{\text{volume (in L)}} = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = 2.85 \times 10^{-3&#10;}

Now, let us assume that mass of C_{12}H_{23}O_{5}N = x grams

And, mass of C_{12}H{22}O_{11} = (1.00 - x)

So, moles of C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}

                              = \frac{x}{369}

Now, moles of C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}

                   = \frac{x}{369} + \frac{(1.00 - x)}{342}

                  = 2.85 \times 10^{-3}

             = x = 0.346

Therefore, we can conclude that amount of C_{12}H_{23}O_{5}N present is 0.346 g  and amount of C_{12}H_{22}O_{11} present is (1 - 0.346) g = 0.654 g.

4 0
3 years ago
Are all the wax rings melting at the same time?
sukhopar [10]

Answer:yes

Explanation:

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