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3 years ago

Suppose you start with 1.80 g of mgco3·5h2o, after complete dehydration how many grams of the sample do you have remaining?

1 answer:

5 0

The number of grams of the sample that remained after hydration is 0.87 grams

calculation
calculate the molar mass of MgCO3. 5 H2O which is

24 + 12+ (3 x16) + ( 5 x18) = 174 g/mol

we well know after hydration water evaporate leaving behind MgCO3
therefore the mass remained is for MgCO3

find the molar mass of MgCO3 = 24 +12 +( 3 x16) = 84 g/mol

the mass remained is therefore

= molar mass of MgCO3 / molar mass of MgCO3.5H2O x 1.80 g

= 84g/mol / 174 g/mol x1.8g = 0.87 grams

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A 25. 00 ml sample of acetic acid containing phenolphthalein indicator is titrated with 0. 1067 m naoh. The solution changes col

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The concentration of acetic acid:

The concentration of the acetic acid before titration is 0.128 M

What is titration?

Titration is a quantitative analytical procedure that works by allowing a known analyte to gradually react with a titrant until an endpoint is reached.

Titration for weak acid and strong base:

Moles of acetic acid = moles of NaOH

Given:

Concentration of NaOH = 0.1067 M

Volume of NaOH = 30.07 ml = 0.03007 L

Calculation:

So, by using the formula, Concentration = Moles/Volume

Moles of NaOH = concentration x volume = 0.1067 x 0.03007 = 0.0032

Therefore, the moles of acetic acid = 0.0032 mole

Now, using the formula again for determining the concentration of acetic acid, we get,

Concentration = Moles/Volume

Concentration of acetic acid = 0.0032/0.025 = 0.128M

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2 years ago

Which one of these elements was suggested as transition element by Mendeleev?

BARSIC [14]

Answer:

The answer is I think None.

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3 years ago

How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s

erastova [34]

Answer : The grams of carbon monoxide needed are 148.89 g

Solution : Given,

Mass of iron, Fe = 198.5 g

Molar mass of iron, Fe = 56 g/mole

Molar mass of carbon monoxide, CO = 28 g/mole

First we have to calculate the moles of iron, Fe.

$\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles$

The balanced chemical reaction is,

$Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)$

From the balanced reaction, we conclude that

2 moles of iron produces from the 3 moles of carbon monoxide

3.545 moles of iron produces from the $\frac{3}{2}\times 3.545=5.3175$ moles of carbon monoxide

Now we have to calculate the mass of carbon monoxide, CO.

$\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}$

$\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g$

Therefore, the grams of carbon monoxide needed are 148.89 g

7 0

3 years ago

How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?

Blababa [14]

Answer:

197mL of 0,506M HCl

Explanation:

The reaction of HCl + BaCO₃ is:

BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.

The moles of BaCO₃ in 9,85 g are:

9,85 g of BaCO₃ × $\frac{1mol}{197,34 g}$ = <em>0,0499 moles of BaCO₃</em>

As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:

0,0499 moles of BaCO₃ × $\frac{2molHCl}{1molBaCO_{3}}$ =<em> 0,0998 moles of HCl</em>

If you have a 0,506M HCl, you need to add:

0,0998 moles of HCl× $\frac{1L}{0,506moles}$ = 0,197 L ≡ 197mL

I hope it helps!

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3 years ago

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihyd

kenny6666 [7]

Answer:

The answer is " $= 0.078 \ kg \ H_2$ ".

Explanation:

calculating the moles in $CH_4 =\frac{PV}{RT}$

$=\frac{(0.58 \ atm) \times (923 \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(232^{\circ} C +273)}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (0.0821 \frac{L \cdot atm}{K \cdot mol})(505)K}\\\\=\frac{(535.34 \ atm \cdot \ L) }{ (41.4605 \frac{L \cdot atm}{mol})}\\\\= 12.9 \ mol$

Eqution:

$CH_4 +H_2O \to 3H_2+ CO \ (g)$

Calculating the amount of $H_2$ produced:

$= 12.9 \ mol CH_4 \times \frac{3 \ mol \ H_2 }{1 \ mol \ CH_4}\times \frac{2.016 g H_2}{1 \ mol \ H_2}\\\\= 78 \ g \ H_2 \\\\= 0.078 \ kg \ H_2$

So, the amount of dihydrogen produced = $0.078 \frac{kg}{s}$

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3 years ago