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Eddi Din [679]
3 years ago
15

Rowena and helga are both performing an experiment with nickel metal. Rowena has a 5 gram sample and determines the density to b

e 8.9g/cm3. If helga has a nickel sample that is twice as large and has a mass of 10 grams what would be the density of helgas sample?
Chemistry
1 answer:
AVprozaik [17]3 years ago
8 0

Answer:

The density of helgas sample is 17.8 g/cm³.

Explanation:

Given that,

Mass of sample of Rowena = 5 gram

Density = 8.9 g/cm³

Mass of sample of helga = 10 gram

We need to calculate the volume of sample

Using formula of volume

V=\dfrac{m}{\rho}

Where, m = mass

\rho = density

Put the value into the formula

V=\dfrac{5}{8.9}

V=0.56\ cm^3

We need to calculate the density of helgas sample

Using formula of density

\rho=\dfrac{m}{V}

Where, m = mass

V = volume

Put the value into the formula

\rho=\dfrac{10}{0.56}

\rho=17.8\ g/cm^3

Hence, The density of helgas sample is 17.8 g/cm³.

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Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s
borishaifa [10]

Answer:

Ca(OH)2 will not precipitate because Q<Ksp

Explanation:

Ksp for Ca(OH)2 has already been stated in the question as 8.0 x 10-8mol2dm-6

The value of the reaction quotient depends heavily on the concentration of the reactants. As the initial concentration of the calcium carbide decreases considerably, the reaction quotient decreases until Q<Ksp hence the Ca(OH)2 will not precipitate from solution.

The reaction equation is:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

From

Ca(OH)2= Ca2+ + 2OH-

Concentration of solution= 0.064×1/64= 1×10-3

Since [Ca2+] = 1×10-3

[OH-]= (2×10-3)^2= 4×10^-6

Hence Q= 4×10^-9

This is less than the Ksp hence the answer.

4 0
2 years ago
(Help would be greatly appreciated) What is the molarity of a solution which contains 22.41 grams of NaCl in 50.0 mL of solution
luda_lava [24]

Answer:

7.67

Explanation:

4 0
3 years ago
13 moles of water to molecules
FromTheMoon [43]
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6 0
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0.5 gm of mixture of NH4Cl and NaCl was boiled with 25 ml of 0.95 N NaOH in a vessel till all the ammonia is expelled.The residu
GalinKa [24]

Answer:

72.66%

Explanation:

NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:

NH₄Cl + NaOH → NaCl + NH₃ + H₂O

The residual NaOH reacts with H₂SO₄ as follows:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

The equivalent-gram of H₂SO₄ are:

16mL * 0.1N * 1.06 = 1.696mEq.

As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:

1.696mEq * (100mL / 10mL) = 16.96mEq.

The mEq of NaOH you add in the first are:

25mL * 0.95mEq = 23.75mEq

That means the NaOH that reacts = moles of NH₄Cl is:

23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =

6.79x10⁻³ moles NH₄Cl

In grams (Using molar mass NH₄Cl = 53.5g/mol):

6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =

0.3633g of NH₄Cl are in the original mixture.

% is:

0.3633g/ 0.5g * 100 = 72.66%

8 0
3 years ago
Noble gas notation write the electron configuration for the manganese atom
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