Ask question

Ask question

All categories

3 years ago

13

What is the main idea and Detail/Evidence of Light microscopes

2 answers:

Vesna [10]3 years ago

7 0

hi I I don’t know

Explanation:

hi

Dimas [21]3 years ago

7 0

The advantage of light microscopes (and stereomicroscopes in particular) is that objects can be looked at with little or no preparation. This makes them very useful for looking at living things, such as flower parts, insects, earthworms and human skin.

You might be interested in

How many molecules are there in 24 grams of Fe? <br><br> (Please explain how you got the answer)

Nataly [62]

1 Answer. SCooke · Stefan V. 1.2×1023 molecules. Hope this helps

6 0

3 years ago

How many liters of Cl2 gas will you have if you are using 63 g of Na?

ELEN [110]

Answer:

You will have 19.9L of Cl2

Explanation:

We can solve this question using:

PV = nRT; V = nRT/P

<em>Where V is the volume of the gas</em>

<em>n the moles of Cl2</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is 273.15K assuming STP conditions</em>

<em>P is 1atm at STP</em>

The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:

63g * (1mol / 70.906g) = 0.8885 moles

Replacing:

V = 0.8885mol*0.082atmL/molK*273.15K/1atm

V = You will have 19.9L of Cl2

6 0

3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

Please someone can help with this answer?

Annette [7]

Answer:

6,000kg/m3

Explanation:

6.00g/1cm3 x 1kg/1000g x 1cm3/0.000001m3

= 6.00kg/0.001m3

= 6,000kg/m3

5 0

3 years ago

0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution what is the molarity

Sunny_sXe [5.5K]

Molarity = moles of solute / liters of solution

M = 0.5 / 0.05

M = 10.0 mol/L⁻¹

hope this helps!

4 0

3 years ago