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1 year ago

Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution?

2 answers:

8 0

Mass = molarity x molar mass( NaNO₃) x volume

mass = 1.50 x 85.00 x 4.50

mass = 573.75 g of NaNO₃

hope this helps!

Fynjy0 [20]1 year ago

4 0

Answer:

D.) 547 g

Explanation:

this is the right answer on edg 2020, just took quiz

hope this helps!!!

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Ammonia has a Kb of 1.8 × 10−5. Find [H3O+], [OH−], pH, and pOH for a 0.310 M ammonia solution.

lawyer [7]

Answer:

[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;

pH = 11.4; pOH = 2.6

Explanation:

The chemical equation is

$\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \,\text{OH}$^{-}$$

For simplicity, let's re-write this as

$\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$$

1. Calculate [OH]⁻

(a) Set up an ICE table.

B + H₂O ⇌ BH⁺ + OH⁻

0.310 0 0

-x +x +x

0.310-x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310$

(b) Solve for [OH⁻]

$\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}$

2. Calculate the pOH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}$

3. Calculate the pH

$\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.6 = \mathbf{11.4}$

4 Calculate [H₃O⁺]

$\text{H$_{3}$O$^{+}$} = 10^{-\text{pH}} = 10^{-11.4} = \mathbf{4.2 \times 10^{-12}} \textbf{ mol/L}$

5 0

1 year ago

How many d electrons (n of dn) are in the central metal ion in the species below? (a) [ru(nh3)5cl]so4 6 d electrons (b) na2[os(c

Georgia [21]

For a) [Ru(NH₃)₅Cl]SO₄
Ru configuration = d⁶s²
In this complex Ru oxidation number is +3
Ru³⁺ configuration = d⁵
number of $d^{n}$ electrons = 5

For b) Na₂[Os(CN)₆]
Os configuration = d⁶s²
In this complex Os oxidation number is +4
Os⁴⁺ configuration = d⁴
number of $d^{n}$ electrons = 4

4 0

1 year ago

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Which of the following energy sources obtains its energy from gravitational potential energy?

ipn [44]

It is a geothermal power plant

6 0

1 year ago

How many moles are in 1.50 g of ethylamine, ch3ch2nh2?

zavuch27 [327]

Mass = mass/molar mass of ch3ch2nh2
the molar mass of CH3CH2NH2 = 12 +(1x3)+12+(1 x2)+14+(1x2) =45 g/mol

moles is therefore = 1.50g /45g/mol = 0.033 moles

6 0

1 year ago

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A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is

musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁ = 1.08atm

V₁ = 250mL

T₁ = 24°C

P₂ = 2.25atm

T₂ = 37.2°C

V₂ = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

T₁ = 24°C, T₁ = 24 + 273 = 297K

T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K

Input the variables and solve for V₂

$\frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}$

V₂ = 124.91mL

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1 year ago