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2 years ago

The elements most likely to form more than type of ion are the....

Chemistry

2 answers:

emmainna [20.7K]2 years ago

7 0

The elements most likely to form more than one type of ion are thetransition metals

Anarel [89]2 years ago

4 0

<span> D transition metals.</span>

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Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in

Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is 2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

$\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\ \frac{m_{S} }{m_{O} }= 2$

Let us express it as

$SO_{\frac{m_{S} }{m_{O} }} = 2$

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

$\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\ \frac{m_{S} }{m_{O} }= 1$

Let us express it as

$SO_2_{\frac{m_{S} }{m_{O} }}= 1$

Now, for the ratio of both the above-calculated ratios,

$\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}$

The required ratio is 2:1

3 0

2 years ago

Determine the mass of MgCl2 needed to create a 100. ml solution with a concentration of 3.00 M.

kiruha [24]

Explanation:

$1000ml \: contain \: 3 \: moles \\ 100 \: ml \: will \: contain \: ( \frac{100 \times 3}{1000} ) \: moles \\ = 0.3 \: moles \\ RFM = 95 \\ 1 \: mole \: weighs \: 95 \: g \\ 0.3 \: moles \: weigh \: ( \frac{(0.3 \times 95)}{1} \: g \\ = 28.5 \: g \: of \: magnesium \: chloride$

5 0

2 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$