During ________________________, a new nuclear membrane forms around eachgroup of chromosomes.

What number of cations is present in 1.17g of sodium chloride ?

taurus [48]

Answer:

1.2 × 10^22 atoms.

Explanation:

Firstly, cations refers to the positively charged atom in the ionic compound, which is Na+.

Given the mass of NaCl as 1.17g, the number of moles of NaCl can be calculated this:

Molar mass of NaCl = 23 + 35.5

= 58.5g/mol

Mole = mass/molar mass

Mole = 1.17/58.5

Mole = 0.02moles

Using Avagadro's number, 6.022 × 10^23 atoms of Na+ are in 1 mole of NaCl.

In 0.02 moles of NaCl, there are 0.02 × 6.022 × 10^23 of Na+

0.1204 × 10^23 atoms

1.2 × 10^22 atoms of Na+ (cation)

5 0

3 years ago

The molar volume of gas at STP, in liters, is?

defon

The molar volume of a gas at STP, in liters, is 22.4

you then multiply 22.4 by 2 to get the answer to the next question:

You can use the molar volume to convert 2 mol of any gas to 44.8 L

You can also use the molar volume to convert 11.2 L of any gas to 0.500 mol.

Avogadro’s law tells you that 1.2 L of O2(g) and 1.2 L of NO2(g) are the same numbers of moles of gas.

7 0

3 years ago

The molar mass of water (H2O) is 18.02 g/mol. Yun needs 0.025 mol H2O for a laboratory experiment. She calculates that she needs

Tomtit [17]

The relation between moles and mass is given by the formula:

# of moles = (mass in grams) / (molar mass)

So, given the molar mass of 18.02 and 0.025 mol, you can solve the formula for the mass of water in grams:

mass of water in grams = # of moles of water * molar mass of water

mass of water in grams = 0.025 mol * 18.02 g / mol = 0.4505 g of water.

Then, she is wrong.

5 0

4 years ago

Read 2 more answers

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08160 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volu

11111nata11111 [884]

Answer:

(a) 13.64; (b) 8.04; (c) 2.25

Explanation:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

$K_{\text{sp}} = {\text{[Ag$^{+}$][I$^{-}$]} = 8.3\times 10^{-17}$

(a) pAg at 35.10 mL

$\text{Moles of I$^{-}$} = \text{0.02500 L} \times \dfrac{\text{0.08160 mol}}{\text{1 L}} = 2.040 \times 10^{-3}\text{ mol/L }\\\text{Moles of Ag$^{+}$} = \text{0.03510 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 1.822 \times 10^{-3}\text{ mol/L}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 1.822 × 10⁻³ 2.040 × 10⁻³

C/mol: -1.822 × 10⁻³ -1.822 × 10⁻³

E/mol: 0 0.218 × 10⁻³

We have a saturated solution of AgI containing 0.218 × 10⁻³ mol of excess I⁻.

V = 25.00 mL + 35.10 mL = 60.10 mL

$\text{[I$^{-}$]} = \dfrac{0.218 \times 10^{-3}\text{ mol}}{\text{0.0610 L}} = 3.57 \times 10^{-3}\text{ mol/L}\\$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s 3.57 × 10⁻³ + s

$K_{\text{sp}} = s(3.57 \times 10^{-3} + s) = 8.3\times 10^{-17}\\$

Check for negligibility:

$\dfrac{3.57 \times 10^{-3}}{8.3\times 10^{-17}} = 4.3 \times 10^{13} > 400\\\\\therefore s \ll 3.63 \times 10^{-3}\\K_{\text{sp}} = s\times 3.63 \times 10^{-3}= 8.3\times 10^{-17}\\\\s = \text{[Ag$^{+}$]} = \dfrac{8.3\times 10^{-17}}{3.63 \times 10^{-3}} =2.29 \times 10^{-14}\\\\\text{pAg} = -\log \left (2.29\times 10^{-14} \right) = \mathbf{13.64}$

(b) At equilibrium

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s s

$K_{\text{sp}} = s\times s = s^{2} = 8.3\times 10^{-17}\\s = \sqrt{8.3\times 10^{-17}} = 9.11 \times 10^{-9}\\\text{pAg} = -\log \left (9.11 \times 10^{-9} \right) = \mathbf{8.04}$

(c) At 47.10 mL

$\text{Moles of Ag$^{+}$} = \text{0.04710 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 2.444 \times 10^{-3}\text{ mol}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 2.444 × 10⁻³ 2.040 × 10⁻³

C/mol: -2.040 × 10⁻³ -2.040 × 10⁻³

E/mol: 0.404 × 10⁻³ 0

V = 25.00 mL + 47.10 mL = 72.10 mL

$\text{[Ag$^{+}$]} = \dfrac{0.404 \times 10^{-3}\text{ mol}}{\text{0.0721 L}} = 5.61 \times 10^{-3}\text{ mol/L}\\\text{pAg} = -\log(5.61 \times 10^{-3}) = \mathbf{2.25}$

6 0

4 years ago

What's wrong with the statement "a mug of hot chocolate contains more heat than a glass of cold water."

frosja888 [35]

Answer:

Heat is not something that can be contained since it is not a tangible thing. Heat is a process

Explanation:

7 0

3 years ago