What is the Calcium iodine cation formula?

Pls help 2-3 tyvm ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎ ‏‏‎ ‎

Triss [41]

Answer:

I could I guess?

But like only 1 lol

Explanation:

6 0

2 years ago

An example of a disaccharide is saccharin-aspartame molecule True False

Flura [38]

I think is False! Because it is not example of a disaccharide it is not saccharin-aspartame molecule.

6 0

2 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago

Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O

Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

$K_{sp}$ = $5.02\times 10^{-6}$

First we have to calculate the solubility of $OH^-$ ion.

The balanced equilibrium reaction will be:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

Let the solubility will be, 's'.

The concentration of $Ca^{2+}$ ion = s

The concentration of $OH^-$ ion = 2s

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][OH^-]^2$

Let the solubility will be, 's'

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=(4s)^3$

Now put the value of $K_[sp}$ in this expression, we get the solubility.

$5.02\times 10^{-6}=(4s)^3$

$s=1.079\times 10^{-2}M$

The concentration of $Ca^{2+}$ ion = s = $1.079\times 10^{-2}M$

The concentration of $OH^-$ ion = 2s = $2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M$

First we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.158\times 10^{-2})$

$pOH=1.67$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33$

Therefore, the pH of a saturated solution is, 12.33

7 0

3 years ago

Identify the type of error in the following scenario. SCENARIO: A group is trying to find the volume of a given liquid. To do th

7nadin3 [17]

Answer:

Error of parallax it usually occurs when the cylinder is above or below the eye level,and thus resulting in differences In reading the millimeters

5 0

2 years ago