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3 years ago

In a chemical compound, would you expect an oxide ion to be joined with one atom of calcium or one atom of potassium? Why?

1 answer:

4 0

I know what you're asking but I don't think the question is stated properly. Technically, an atom will not join with an "oxide" ion; i.e., the oxide ion is an atom of oxygen to which two electrons have been added. An oxide ion will add to 2 K ions or 1 Ca ion. The K ion has lost just one electron so it takes two of them to equal the 2- charge on the oxide ion whereas the Ca ion has lost two electrons and it takes only one of them to equal the charge on the oxide ion.

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A gas occupies a volume of 1.00 L at 25.0°C. What volume will the gas occupy at 1.00 x10^2 °C?

Leno4ka [110]

Answer : The volume of gas occupy at $1.00\times 10^2^oC$ is, 1.25 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K$

Putting values in above equation, we get:

$\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L$

Therefore, the volume of gas occupy at $1.00\times 10^2^oC$ is, 1.25 L

3 0

3 years ago

Consider the reaction 3X + 2Y 5C + 4D

xeze [42]

Answer:

This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the molesof X that are needed to

produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:Mole ratios:In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.

Calculations:In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:

Explanation:

mark me brainliest!!

5 0

2 years ago

When a skin diver is on the top of the water at 1.03 atm, her lung volume is 3.62L. How will the volume of her lungs change as s

Anon25 [30]

Answer:

–2.23 L

Explanation:

We'll begin by calculating the final volume. This can be obtained as follow:

Initial pressure (P₁) = 1.03 atm

Initial volume (V₁) = 3.62 L

Final pressure (P₂) = 2.68 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

1.03 × 3.62 = 2.68 × V₂

3.7286 = 2.68 × V₂

Divide both side by 2.68

V₂ = 3.7286 / 2.68

V₂ = 1.39 L

Finally, we shall determine the change in volume. This can be obtained as follow:

Initial volume (V₁) = 3.62 L

Final volume (V₂) = 1.39 L

Change in volume (ΔV) =?

ΔV = V₂ – V₁

ΔV = 1.39 – 3.62

ΔV = –2.23 L

Thus, the change in the volume of her lung is –2.23 L.

NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!

3 0

2 years ago

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

5 0

2 years ago

4. If the DNA nitrogen bases were TACCGGAT, how would the other half of

Elena L [17]

Answer:

ATGGCCTA

Explanation:

For this we have to keep in mind that we have a specific relationship between the nitrogen bases:

-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.

-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.

Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:

TACCGGAT

ATGGCCTA

I hope it helps!

6 0

2 years ago