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3 years ago

A solution is made by mixing 15.0 g of sr(oh)2 and 55.0 ml of 0.200 m hno3.

Chemistry

1 answer:

noname [10]3 years ago

4 0

Letter A: The balanced equation is:

Sr(OH)2 (s) + 2 HNO3 (aq) → Sr(NO3)2 (aq) + 2 H2O

Letter B: The solution is:

(15.0 g Sr(OH)2) / (121.6358 g Sr(OH)2/mol) = 0.12332 mol Sr(OH)2
(0.0550 L) x (0.200 mol/L HNO3) = 0.011 mol HNO3

0.011 mole of HNO3 would respond entirely with 0.011 x (1/2) = 0.0055 mole of Sr(OH)2 but there is more Sr(OH)2 existing than that, so Sr(OH)2 is in excess and HNO3 is the limiting reactant.

( 0.011 mol HNO3) x (1/2) = 0.0055 mol Sr(NO3)2

Since the NO3{-} ions continued in solution during, the concentration of the NO3{-} ions didn't change during the reaction, so they are still 0.200 M.
The concentration of Sr{+2} ions is (0.0055 mol Sr(NO3)2) / (55.0 mL) = 0.100 M
Since HNO3 is the limiting reactant, no H{+} ions continue at the end of the reaction.

Letter C: Since all of the acid reacted and there was still Sr(OH)2 left over, the resulting solution would be basic.

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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately <u>333 grams.</u>

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2 years ago

Why calcium oxide and magnesium oxide used as soil treatment

BartSMP [9]

Answer:

Chemically speaking, lime refers only to calcium oxide

(CaO); however, in common usage the term includes the calcination products of calcitic and dolomitic limestones. Calcitic (high-calcium) limes are produced by calcination of

calcareous materials (e.g., calcitic limestone, calcite,

oyster shells, and chalk) containing from 95 to 99 percent

calcium carbonate (CaCO^). Dolomitic limes are produced from

dolomitic limestone or dolomite which contains from 30 to ^+0

percent magnesium carbonate (MgCO^), the rest being calcium

carbonate.

At atmospheric pressure, calcite in limestone decomposes

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form MgO, CO^ and CaCO^. It is necessary to raise the temperature to 900°C to decompose the CaCO^ (15, 35)» This

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Various investigators have studied the effects of stone

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Explanation:

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3 years ago

1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)

Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of carbon to fore 0.3009 moles of CO₂

The maximum mass of carbon dioxide that can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

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3 years ago

A chemistry teacher is able to grade chemistry labs at a rate of 5 labs for every 3 minutes. They need to grade 165 labs how man

PIT_PIT [208]

Answer:

1.65hr

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