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2 years ago

What is empirical formula?

Chemistry

2 answers:

Ulleksa [173]2 years ago

6 0

Answer:

the empirical formula of a chemical compound is the simplest whole number ratio of atoms present in a compound.

from the image

N2O4 will be NO2

P4O10 will be P2O5

and so on

Explanation:

brainliest plz

Evgen [1.6K]2 years ago

6 0

a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

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What is the energy of a radio wave whose frequency is 6.2 x 106 Hz?

slava [35]

Answer:

657.2 Hz

Explanation:

7 0

2 years ago

Find the linear function Cequalsf(F) that gives the reading on the Celsius temperature scale corresponding to a reading on the

padilas [110]

Answer:

C = (5/9) F - (160/9)

They both read equal at Z = - 40

Explanation:

We are looking for a linear function so we can write the following condition

Y = aX + b

Applying it to the exercise we got C = a F + b

Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212

0 = 32 a + b (1)

100 = 212 a + b (2)

From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)

and b = - (5/9)32 = - 160/9

Finally the linear function is C = (5/9) F - (160/9)

Both readings are equal at a Z number so

Z = (5/9) Z - 160/9

(4/9) Z = -160/9 and Z = - 40

6 0

3 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago

22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass

Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0

3 years ago

A person’s blood alcohol (C2H5OH) level can be determined by titrating a sample of blood plasma

Elanso [62]

Answer:

0.17%

Explanation:

With the equation:

2Cr2O7 2- + C2H5OH + H2O --> 4Cr3+ + 2CO2 + 11H2O

We can assume that every mole of ethanol needs 2 moles of Dichromate to react.

So if in 1L we have 0.05961 moles of dichromate we can discover how many moles we have in 35.46mL

1000 mL - 0.05962 moles

35.46 mL - x

x = $\frac{0.05962 * 35.46}{1000}$

x = 2,11* 10^-3 moles

As we said earlier, 1 mole of ethanol needs 2 mole of dichromate, so in the solution we have 1,055*10^-3 moles of ethanol. We can discover the mass of ethanol present in the solution.

1 mole - 46g

1.055*10^-3 - y

y = 46 * 1.055*10^-3

y = 0.048 g

To discover the percent of alchol we can use a simple relation

28 g - 100%

0.048 - z

z = $\frac{0.048 * 100}{28}$

z = 0.17%

6 0

3 years ago

What is empirical formula?​

What is empirical formula?