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2 years ago

Xxxx only girls can join gca-vtki-tqh

Chemistry

2 answers:

4vir4ik [10]2 years ago

6 0

Answer:

it is the app for homework not for nonsense things.

do good be good and see good and think good

ValentinkaMS [17]2 years ago

3 0

Answer:

ok ok i'll search it

Explanation:

btw thanks for the points and don't waste it!

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Which compound is made from metals and non metals

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It’s -> Ionic compound

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3 years ago

How many atoms are present in 65.39 g of zinc?

klio [65]

<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

65.39 g Zn ( 1 mol / 65.38 g ) ( </span>6.022 x 10^23 atoms / 1 mol ) = 6.023x10^23 atoms Zn

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3 years ago

Give five properties of water.

Gelneren [198K]

Polarity, cohesion, adhesion, surface tension, high specific heat, and evaporating cooling

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2 years ago

Read 2 more answers

Will mark the brainiest later for correct answers! Please show work.

Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}$

a. moles in 14.08 g of $C_{12}H_{22}O_{11}$ = $\frac{14.08g}{342.3g/mol}=0.04113moles$

molecules in 14.08 g of $C_{12}H_{22}O_{11}$ = $0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}$

b. moles in 17.75 g of NaCl = $\frac{17.75g}{58.5g/mol}=0.3034moles$

molecules in 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

formula units 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

c. moles in 20.06 g of $CuSO_4.5H_2O$ = $\frac{20.06g}{249.68g/mol}=0.08034moles$

formula units in 20.06 g of $CuSO_4.5H_2O$ = $0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}$

7 0

3 years ago

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a

natali 33 [55]

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

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

We are given that, $K_{f} = 1.0 \times 10^{31}$

and, $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

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

K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

= $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

4 0

2 years ago

Xxxx only girls can join gca-vtki-tqh ​

Xxxx only girls can join gca-vtki-tqh