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Wewaii [24]
2 years ago
15

A solution of sucrose (sugar) in water is in equilibrium with solid sucrose. If more solid sucrose is now added, with stirring,

Chemistry
1 answer:
denis-greek [22]2 years ago
7 0

When equilibrium is reached, the solution is said to be saturated. A solution containing a higher concentration of solute than its solubility is said to be supersaturated.

Answer: d

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8. Balance the following nuclear equations:<br> ^65/30Ca -&gt; ^65/29Sc +_______
a_sh-v [17]

Explanation:

          ⁶⁵₃₀Ca   →   ⁶⁵₂₉Sc  +   ⁿₓH

The reaction above is nuclear reaction.

In a nuclear reaction, the mass number and atomic number must be conserved.

The mass number is the superscript before the atom

Atomic number is the subscript before the atom

  Conserving mass number:

       65 = 65 + n

        n = 0

   conserving atomic number:

    30 = 29 + x

     x = 1

The unknown atom is a positron i.e a positively charged electron:   ⁰₁e

              ⁶⁵₃₀Ca   →   ⁶⁵₂₉Sc  +  ⁰₁e

learn more:

Transmutation brainly.com/question/3433940

#learnwithBrainly

3 0
3 years ago
Given that a for HCN is 6. 2×10^−10 at 25 °C. What is the value of b for cn− at 25 °C?
Kay [80]

If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).

<h3>What is base dissociation constant? </h3><h3 />

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 6.2× 10^(-10)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{6.2×10^(-10) }

= 1.6× 10^(-5)

Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).

learn more about base dissociation constant :

brainly.com/question/9234362

#SPJ4

4 0
2 years ago
Find the volume of an object that has a density of 3.14 g/mL and a mass of 52.9 g.
Natasha_Volkova [10]

Answer:

The answer is

<h2>16.8 mL</h2>

Explanation:

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

volume =  \frac{mass}{density}  \\

From the question

mass of object = 52.9 g

density = 3.14 g/mL

The volume of the object is

volume =  \frac{52.9}{3.14}  \\   = 16.847133757...

We have the final answer as

<h3>16.8 mL</h3>

Hope this helps you

5 0
3 years ago
SnO2 + 2 H2 ——&gt; Sn + 2 H2O
SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂  = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

             SnO₂   +    2H₂    →   Sn  +   2H₂O

Ensure that the equation given is balanced;

       

Now,

          the known species is water;

                  6.02 x 10²³ molecules of water  = 1 mole

                   1.2 x 10²¹ molecules of water  = \frac{1.2 x 10^{21} }{6.02 x 10^{23} }    = 0.2 x 10⁻²moles

Number of moles of water  = 0.002moles

           From the balanced chemical equation:

         

             2 mole of water is produced from 1 mole of    SnO₂  

           0.002 moles of water will be produced from \frac{0.002}{2}  = 0.001moles

To find the mass;

           Mass  = number of moles x molar mass

Molar mass of  SnO₂ = 118.7 + 2(16) = 150.7g/mol

        Mass  =  0.001 x 150.7 = 0.15g

3 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
3 years ago
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