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3 years ago

Please I really need help with this

Chemistry

2 answers:

sesenic [268]3 years ago

5 0

Answer:

B. It is used to open and close the circuit.

Ksju [112]3 years ago

3 0

Answer:

B. it is used to open and close the circuit

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Which compound contains both ionic and covalent

DedPeter [7]

(1) CaCO3 is your answer. The positive ion is $\text{Ca}^{2+$ and the negative radical ion is $\text{CO}_3^{-}$ . There exists covalent bonds between the carbon and oxygen atoms in the $\text{CO}_3^{-}$ .

6 0

4 years ago

How many mL of 8.00M HCl solution are needed to make a dilution with a final volume of 2,500.ml, and a

kompoz [17]

Answer:

104.1 mL

Explanation:

c1V1 = c2V2

V1 = c2V2/c1 = (0.333 M×2500 mL)/8.00 M

V1 = 104.1 mL

5 0

3 years ago

Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years

Sliva [168]

Answer : The time passed in years is $2.83\times 10^3\text{ years}$

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

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

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

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

t = time passed by the sample = ?

a = let initial amount of the reactant = X g

a - x = amount left after decay process = $71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg$

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}$

$t=2831.00\text{ years}=2.83\times 10^3\text{ years}$

Therefore, the time passed in years is $2.83\times 10^3\text{ years}$

8 0

3 years ago

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

Gre4nikov [31]

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

$\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M$

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially conc. 0 0 0.1576

At eqm. (x) (x) (0.1576-2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}$

By solving the term, we get:

$x=0.0742,0.0839$

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

5 0

4 years ago

How many grams of chlorine gas can be produced if 15 grams of FeCl3 reacts with 4 moles of O2? What is the limiting reactant? Wh

Mariana [72]

Answer:

The limiting reactant is FeCl3

The excess reactant is O2

The theoretical yield Cl2 is 9.84 grams

The % yield = 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles of O2 = 4.0 moles

Mass of Cl2 = 9.5 grams = actual yield

Step 2: The balanced equation

4 FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate the limiting reactant

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

FeCl3 has the smallest amount of moles, this is the limiting reactant. It will be completely consumed ( 0.0925 moles).

O2 is in excess. There will react 3/4 * 0.0925 = 0.0694 moles

There will remain 4.0 - 0.0694 = 3.3904 moles O2

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 we'll have 6/4 * 0.0925 = 0.13875 moles Cl2

Step 6: Calculate mass of Cl2

Mass Cl2 = moles Cl2 * molar mass Cl2

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams = theoretical yield

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) *100%

% yield = (9.5 grams / 9.84 grams) * 100%

% yield = 96.5 %

3 0

4 years ago