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3 years ago

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Select the true statement. ANSWER Unselected Anions are smaller than their neutral parent atoms because there are fewer electron

s “competing” for the attractive force of the protons in the nucleus. Unselected Anions are smaller than their neutral parent atoms because there are more electrons “competing” for the attractive force of the protons in the nucleus. Unselected Cations are smaller than their neutral parent atoms because there are fewer electrons “competing” for the attractive force of the protons in the nucleus. Unselected Cations are smaller than their neutral parent atoms because there are more electrons “competing” for the attractive force of the protons in the nucleus. Unselected

1 answer:

Olenka [21]3 years ago

3 0

Answer:

Explanation:

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How many moles of aluminum are in 4.35 moles of al2si4o10(OH)2

Gnoma [55]

Answer:

8.7 mol Al

Explanation:

You have the compound Al₂Si₄O₁₀(OH)₂. Within the compound, you have atoms. You can tell how many atoms there are by looking at the subscript. You have 2 aluminum atoms, giving you 8.7 moles of aluminum.

4.35 × 2 = 8.7

6 0

3 years ago

What is one of the products produced when al(no3)3 and cao react together? alo2 cano3 al2o3 ca(no3)3?

cricket20 [7]

The chemical balanced equation for the given question is this:
2AL[NO3]3 + 3CaO = AL2O3 + 3Ca[NO3]2.
So, one of the product formed is AI2O3. The other product is Ca[NO3]2.
The reaction is a double replacement reaction.

8 0

3 years ago

Read 2 more answers

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

1 Calcium carbonate breaks down on heating to produce calcium oxide and carbon dioxide gas.

Alborosie

Answer: See below

Explanation:

1. a) 0.15 moles calcium carbonate (15g/100g/mole)

b) 0.15 moles CaO (molar ratio of CaO to CaCO3 is 1:1)

c) 8.4 grams CaO (0.15 moles)*(56 grams/mole)

2. a) 0.274 moles Na2O (17g/62 grams/mole)

b) 46.6 grams NaNO3 (2 moles NaNO3/1 mole Na2O)*(0.274 moles Na2O)*(85 g/mole NaNO3)

7 0

3 years ago

There is a 30g of be-11 it has a half-life of about 14 seconds how much will be left in 28 seconds

Lelu [443]

There will be 7.5 g of Be-11 remaining after 28 s.

If 14 s = 1 half-life, 28 s = 2 half-lives.

After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.

After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.

In symbols,

<em>N</em> = <em>N</em>₀(½)^<em>n</em>

where

<em>n</em> = the number of half-lives

<em>N</em>₀ = the original amount

<em>N</em> = the amount remaining after <em>n</em> half-lives

6 0

3 years ago