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2 years ago

What is the stable ion of beryllium.

Chemistry

2 answers:

Oxana [17]2 years ago

7 0

Answer:

it is deez

Explanation:

8uyygl.huyiuo9j

Archy [21]2 years ago

4 0

Answer:

Be^2+ ion

Explanation:

Be(OH)x^y- ions and Be(OH)2 solid

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Pls, Help ASAP with 30 points!!!

Lapatulllka [165]

Answer:

the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy

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2 years ago

Write Qc for each of the following:(1) Gaseous sulfur tetrafluoride reacts with liquid water to produce gaseous sulfur dioxide a

snow_tiger [21]

Answer : The expression for reaction quotient will be :

(1) $Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) $Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

Explanation :

Reaction quotient $(Q_c)$ : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

(1) The given balanced chemical reaction is,

$SF_4(g)+2H_2O(l)\rightarrow SO_2(g)+4HF(g)$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :

$Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) The given balanced chemical reaction is,

$2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

7 0

2 years ago

The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days

professor190 [17]

Answer:

6.2 g

Explanation:

In a first-order decay, the formula for the amount remaining after n half-lives is

$N = \frac{N_{0}}{2^{n}}$

where

N₀ and N are the initial and final amounts of the substance

1. Calculate the number of half-lives.

If $t_{\frac{1}{2}} = \text{3.8 da}$

$n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}$

2. Calculate the final mass of the substance.

$\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}$

4 0

2 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

Explain, in terms of energy, why the temperature of the reaction mixture increases

Dima020 [189]

Answer:

Magnesium reacts with dilute hydrochloric acid in a conical flask which is connected to an inverted measuring cylinder in a trough of water. The volume of hydrogen gas produced is measured over a few minutes, and the results are used to plot a graph........

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2 years ago