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2 years ago

How many atoms is 2.34 moles of titanium?

Chemistry

2 answers:

uysha [10]2 years ago

8 0

$1.409 \times {10}^{24} \: \: atoms$

Explanation:

Convert from moles to atoms:

2.34 mol titanium (Ti)

Determine the amount of substance, n:

n = 2.34 mol

Look up Avogadro's constant, $N_A$ , to find the number of atoms in a mole:

$N_A$ = 6.022×10^23 atoms/mol

Multiply n by $N_A$ to convert to the number of atoms, $N_{atoms}$ :

Precise Answer:

$N_{atoms}$ = n× $N_A$ = (2.34 mol)/1×(6.022×10^23 atoms)/(1 mol) = 1.409×10^24 atoms

BartSMP [9]2 years ago

7 0

2.34 moles titanium x (6.022 x 10^23)/1 mole titanium = 1.41 x 10^24

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Calcium carbonate has the chemical formula CaCO3. A certain quantity of calcium carbonate has 8.35x1025 atoms. How many moles of

kotegsom [21]

Answer

27.7 moles

Explanation:

Data Given:

Chemical formula of Calcium carbonate = CaCO₃

no. of atoms of CaCO₃ = 8.35 x 10²⁵ atoms

moles of CaCO₃= ?

Solution

First convert number of atoms to number of molecules

we have 5 atoms per molecule then how many molecules will be for 8.35 x 10²⁵ atoms

So, apply unity formula

5 atoms ≅ 1 molecule of CaCO₃

8.35 x 10²⁵ atoms ≅ X molecule of CaCO₃

by doing cross multiplication

X molecule of CaCO₃ = 8.35 x 10²⁵ atoms x 1 molecule / 5 atoms

X molecule of CaCO₃ = 1.67 x 10²⁵ molecule

Formula will be used

no. of moles = no. of molecules / Avogadro's number . . . . (1)

Where

Avogadro's number = 6.022 x 10²³ molecules/mol

So,

Put values in equation 1

no. of mole = 1.67 x 10²⁵/ 6.022 x 10²³ (molecules/mol)

no. of mole = 27.7 moles

So,

moles of CaCO₃ = 27.7 moles

5 0

3 years ago

Many important biochemicals are organic acids, such as pyruvic acid ( p K a = 2.50 ) and lactic acid ( p K a = 3.86 ) . The conj

Ivenika [448]

Answer:

Pyruvic acid: conjugate base

Lactic acid: conjugate base

Explanation:

The ratio of conjugate base to conjugate acid can be found using the Henderson-Hasselbalch equation when the pH and pKa are known.

pH = pKa + log([A⁻]/[HA])

The equation can be rearranged to solve for the ratio:

pH - pKa = log([A⁻]/[HA])

[A⁻]/[HA] = 10^(pH-pKa)

Now we can calculate the ratio for the pyruvic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 2.50) = 79433

[A⁻] = 79433[HA]

There is a much higher concentration of the conjugate base.

Similarly for lactic acid:

[A⁻]/[HA] = 10^(pH-pKa) = 10^(7.4 - 3.86) = 3467

[A⁻] = 3467[HA]

For lactic acid the conjugate base also dominates at pH 7.4

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4 years ago

PLEASE HELP ME TAKE SOO LONG! (ima fail this test!) its a study guide!

kakasveta [241]

I think it might be C or D

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3 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

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3 years ago

Which mineral group dose copper belong

RideAnS [48]

Answer:

Native Elements; Metallic Elements

Explanation:

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3 years ago