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3 years ago

How much Sr(OH)2 • 8 H2O (M = 265.76) is needed to prepare 250.0 ML of solution in which [OH]= 0.100M?

2 answers:

6 0

Use the formula
first step:
Use the formula
molarity= mole/liter
change ml to l
plug in data
to get .1=mole/.25 or .1M*.25liter
which =.025 moles
then divide .025 moles by two because there are two OH in Sr(OH)2
then multiply that by 265.76 (the molar mass of water)
.0125*265.76
which is 3.32grams this is your answer

Flura [38]3 years ago

5 0

The answer is = 3.322grams

The explanation:

1-First we need o get the no.of moles:

moles = molarity * volume(L)

by substitution:

∴ moles = 0.1M * 0.25 L

= 0.025 moles

but we have 2 OH in Sr(OH)2

so moles = 0.025 /2 = 0.0125 moles

2- then we need to get how many grams of it so, we are going to use this formula:

mass = moles * molar mass

by substitution:

∴ mass = 0.0125 * 265.76

= 3.322 grams

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Which water usage uses the least amount of water in a year in the United States

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Answer: a

Explanation:

Industry uses only about 18% while the others use around 70-90% of water.

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2 years ago

Predict the sign of the entropy change,Delta S, for each of the following reactions:The signs are either going to be pos or nega

algol [13]

Answer: a) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$ : negative

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$ : positive

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$ : positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$ : negative

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$ : positive.

f) $I_2(s)\rightarrow I_2(g)$ : positive.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) $Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)$

As ions are moving to solid form , randomness decreases and thus sign of $\Delta S$ is negative.

b) $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

c) $2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As 2 moles of reactants are converted to 4 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

d) $P_4(g)+5O_2(g)\rightarrow P_4O_{10}(s)$

As gas is changing to solid, randomness decreases and thus sign of $\Delta S$ is negative.

e) $C_4H_8(g)+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)$

As 7 moles of reactants are converted to 8 moles of products , randomness increases and thus sign of $\Delta S$ is positive.

f) $I_2(s)\rightarrow I_2(g)$

As solid is changing to gas, randomness increases and thus sign of $\Delta S$ is positive.

6 0

3 years ago

7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?

Illusion [34]

Answer:

7. 0.1021 M

8. 1.167 M

10. Increase in volume of water would lower the rate of reaction

Explanation:

7. What is the molar concentration of H₂C₂O₄ ?

Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.

Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V

= 1.225 × 10⁻³ mol/12 × 10⁻³ L

= 0.1021 mol/L

= 0.1021 M

8. Using the data from question 7 what is the molar concentration of KMnO₄ ?

Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.

Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L

So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V

= 14 × 10⁻³ mol/12 × 10⁻³ L

= 1.167 mol/L

= 1.167 M

10. From question number 7, what effect increasing the volume of water has on the reaction rate?

Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.

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3 years ago

Select the correct answer. The gas in a sealed container has an absolute pressure of 9.25 atmospheres. If the air around the con

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Answer:

Explanation:

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3 years ago

Read 2 more answers

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Bingel [31]

Answer:

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Explanation:

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4. 2 compounds become one compound

5. ca is replaced with na and na is replaced with ca

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3 years ago