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2 years ago

If 23.17 g of beryllium (Be) reacts with water at standard temperature and pressure,

Chemistry

1 answer:

emmainna [20.7K]2 years ago

5 0

Volume of H2 produced = 57.6576 L

<h3>Further explanation</h3>

Given

23.17 g Be

Required

Volume of H2

Solution

Reaction

Be(s)+H2O(g)→BeO(s)+H2(g)

mol Be :

= 23.17 g : 9 g/mol

= 2.574

From the equation, mol H2 : mol Be = 1 : 1, so mol H2 = 2.574

Volume H2(assumed at STP, 1 mol=22.4 L) :

= 2.574 x 22.4 L

= 57.6576 L

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Assuming your weight is 440.55 N and the force of gravity on Earth is 9.8 m/s2 what is your mass? *

Alexandra [31]

Answer:

44.95408163 kg

44.95 kg (2 decimal places)

Explanation:

Weight=Mass*Gravity

440.55=m*9.8

Make m the subject

M=440.55/9.8

M= 44.95 kg

5 0

3 years ago

How many moles are in 1.20 times 10^25 atoms of phosphorus

Stels [109]

19.927 moles are in 1.20 times $10^{25}$ atoms of phosphorus.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others.

1 mole of any substance contain Avogadro's number of molecules so we can calculate the number of moles by dividing the provided number of atoms over Avogadro's number to obtain the number of moles .

Moles= $\frac{Atoms}{\;Avogadro's \;number }$

Moles= 1.20 X $10^{25}$ atoms ÷ 6.022 X $10^{23}$

= 19.927

Hence, 19.927 moles are in 1.20 times $10^{25}$ atoms of phosphorus.

Learn more about moles here:

brainly.com/question/26416088

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5 0

1 year ago

The temperature of evaporation is much higher for water than for alcohol. Without knowing more about the chemistry of alcohol, w

3241004551 [841]

Answer:

Explanation:

Alcohols are organic molecules characterized majorly by the presence of the OH group in their molecule. The OH group is majorly responsible for several of their characteristics. This include the formation of hydrogen bonds between alcohol molecules. While this makes them more inorganic than most organic compounds, comparatively the hydrogen bonding formed in alcohols is not as strong as that which is present in water.

The higher strength of the hydrogen bonding is responsible for some comparable properties. While water boils at a temperature of 100 degrees Celsius, alcohol boils at a temperature of 78 degrees Celsius. This is an evidence to the fact that hydrogen bonding in alcohol is less stronger that that in water.

4 0

3 years ago

How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃? *

densk [106]

Answer:

26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃

Explanation:

To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?

$molesofO_{2} =\frac{35 moles of FeCl_{3}*3 moles of O_{2} }{4 moles of FeCl_{3}}$

moles of O₂= 26.25 ≅ 26.3

<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>

7 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago