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aleksandr82 [10.1K]

3 years ago

8

24.08 x 1023 atoms of boron (b) is equal to how many moles of boron? 5 b 10.81 boron a. 2 moles b. 4 moles c. 12 moles d. 6.02 x

1023 moles

2 answers:

bija089 [108]3 years ago

7 0

24.08 x 10²³<span> atoms of boron (b) is equal to how many moles of boron?
</span>
24,08×10²³ / 6,02×10²³=4

<span>b. 4 moles</span>

san4es73 [151]3 years ago

6 0

The answer would be D!

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Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g

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Answer:

$\large \boxed{84.7 \, \%}$

Explanation:

Mᵣ: 58.44 278.11

Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g: 26.3

1. Moles of NaCl

$\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}$

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$\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}$

(c) Theoretical yield of PbCl₂

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(d) Percent yield

$\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}$

7 0

3 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

5 0

3 years ago

if the density of water is 1 g/ml, what would happen to an object with a mass of 80 g and a volume of 100 cubic cm when you plac

Nastasia [14]

Answer:

<em>that</em><em> </em><em>obj</em><em>ect</em><em> </em><em>will</em><em> </em><em>float</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>water</em>

Explanation:

Since the density of water is greater than that of the object , the object will float on the water.

NB. Density of the object=

$\frac{mass}{volume}$

Given that mass of the object is 80g and volume 100cm^3

substituting them in the formula

Density=80g/100cm^3

Density=0.8g/cm^3

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5 0

3 years ago

This scientist proposed that atoms have a nucleus.

Alik [6]

Ernest Rutherford is the answer.

Hope this helps!!

4 0

3 years ago