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3 years ago

Problem:

Chemistry

1 answer:

adelina 88 [10]3 years ago

7 0

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

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Explain why mineral oil which is a mixture of hydrocarbons does not dissolve in water

Mkey [24]

Oil molecules like to stick to other oil molecules more than they like to stick to water molecules. This is because mineral oil has nonpolar molecules and water has polar ones.

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3 years ago

For the reaction below , label each reactant as an electron pair acceptor or electron pair donor and as a Lewis acid or a Lewis

JulsSmile [24]

Since the reaction shown in the question is an acid - base reaction in the Lewis sense; the Lewis acid here is AlCl3 while the Lewis base here is Cl^- .

<h3>What is a Lewis acid?</h3>

A Lewis acid is a substance that accepts electron pair while a Lewis base donates an electron pair.

Now consider the given reaction; AlCl3 +Cl^- ------> AlCl 4 ^-. The Lewis acid here is AlCl3 while the Lewis base here is Cl^- .

Learn more about acid - base reaction: brainly.com/question/14356798

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2 years ago

Consider this initial-rate data at a certain temperature for the reaction described by N2O3(

Yuliya22 [10]

N2O3(g)→NO(g)+NO2(g)

[N2O3] Initial Rate
0.1 M r(t)=0.66 M/s
0.2 M r(t)=1.32 M/s
0.3 M r(t)=1.98 M/s

We can have the relationship:

([N2O3]/[N2O3]0)^m=r(t)/r0(t)
However,
([N2O3]/[N2O3]0) = 2

Also, we assume m=1 which is the order of the reaction.

Thus, the relationship is simplified to,

r(t)/r0(t) = 2

r(t)=k[N2O3]

0.66 M/s=k×0.1 M

k=6.6 s−1

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3 years ago

Read 2 more answers

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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3 years ago

Explain the reaction between metallic hydroxides and a dilute acid?

dolphi86 [110]

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