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3 years ago

Which property of chlorine explains why it is likely to form a compound with an alkali metal

Chemistry

1 answer:

Aleksandr [31]3 years ago

4 0

Answer:

a. It will accept one electron to complete its outer shell.

Explanation:

Chlorine is in Group 17, so it has seven valence electrons, one short of a complete outer shell.

Thus, it will accept an electron from sodium to complete its outer shell.

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Determine the empirical formula for succinic acid that is composed of 40.60% carbon, 5.18% hydrogen, and 54.22% oxygen.

aleksklad [387]

Answer:

C₂H₂O₃

Explanation:

The empirical formula of a compound is derived bu finding the whole ratios of the constituent elements.

In succinic acid, the ratios of carbon to hydrogen to oxygen is calculated as follows:

% mass

Carbon- 40.60

Hydrogen - 5.18

Oxygen - 54.22

RAM

Carbon -12

Oxygen - 15.994

Hydrogen -1.008

No of moles elements in the compound

Carbon = 40.60/12=3.3833

Oxygen = 54.22/15.994= 3.39

Hydrogen= 5.18/1.008 = 5.1389

Mole ratios of the individual elements we divide by the smallest value of the number of moles.

Carbon: Hydrogen : Oxygen

3.3833/3.3833:3.39/3.3833:5.1389/3.3833

=1:1:1.5

We can multiply the value by 2 to get the whole number ratio.

=2:2:3

The empirical formula will be:

C₂H₂O₃

5 0

3 years ago

Consider the following reaction where Kc = 1.80×10-2 at 698 K:2 HI (g) H2 (g) + I2 (g)A reaction mixture was found to contain 0.

Iteru [2.4K]

Answer:

Answer for the given statements: (1) T , (2) F , (3) T , (4) F , (5) F

Explanation:

At the given interval, concentration of HI = $\frac{0.304}{1}M=0.304M$

Concentration of $H_{2}$ = $\frac{5.07\times 10^{-2}}{1}M=5.07\times 10^{-2}M$

Concentration of $I_{2}$ = $\frac{4.57\times 10^{-2}}{1}M=4.57\times 10^{-2}M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[H_{2}][I_{2}]}{[HI]^{2}}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(5.07\times 10^{-2})\times (4.57\times 10^{-2})}{(0.304)^{2}}=2.51\times 10^{-2}$

So, the reaction is not at equilibrium.

As $Q_{c}> K_{c}$ therefore reaction must run in reverse direction to reduce $Q_{c}$ and make it equal to $K_{c}$ . That means HI(g) must be produced and $H_{2}$ must be consumed.

3 0

2 years ago

Plz help me with this problem

Ilya [14]

Answer:

4feet

Explanation:

Because when he moves he only goes 4 miles every time I

6 0

3 years ago

Read 2 more answers

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

KIM [24]

Answer: The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5730 years

Putting values in above equation, we get:

$k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 25) = 75 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs$

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0

3 years ago

WILL GIVE BRAINLYST

Andrew [12]

Answer: 12.18 u

Explanation: The average atomic mass of an element is calculated by taking the weighted average of the atomic masses of its stable isotopes.

In other words, each stable isotope will contribute to the average mass of the element proportionally to its abundance.

3 0

2 years ago