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katrin2010 [14]
2 years ago
15

The table shows the densities of four samples, each of the same volume.

Chemistry
2 answers:
omeli [17]2 years ago
5 0

Answer:

Its Y

Explanation:

7nadin3 [17]2 years ago
4 0
The formula for density is D=m/v therefore the formula for mass is m=d(v) since v is a constant the sample with the highest density will also have the highest mass. Meaning that Y is the answer.
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What is a characteristic of all fuel cells?
CaHeK987 [17]
B. Electrical energy is produced from oxidation reactions. 
I don't have an explanation for this though. Do you need one? I can probably look it up.
8 0
2 years ago
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A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
Leona [35]

Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles

0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
Tamiku [17]

Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow  Na_2SO_4(aq) + CO_2(g) + H_2O(l)

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

5 0
2 years ago
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam
mafiozo [28]

Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

A. 0.272 mol

B. 0.04 mol

C. 0.997 mol

D. 0.139 mol

E. 0.0681 mol

Answer:

The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

Vapor pressure of the liquid = 18 torr = 2399.803 Pa

Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm  

= 100007.775 Pa

Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa

Volume of H₂ = 1.7 L = 0.0017 m³

Temperature = 20 °C = 293.15 K

Therefore,

n = \frac{PV}{RT} =  \frac{100007.775 \times 0.0017 }{8.3145 \times 293.15} = 0.068078 \ moles

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.

7 0
2 years ago
An unknown metal is dropped into 127 grams of water. The temperature of the water has been raised from 25OC to 28OC. Using the s
Gnesinka [82]

Answer:

he amount of heat gained by the water is 1.59 kJ    

Explanation:

Relation between heat energy, specific heat and temperature change is as follows

Q = mCΔT

where,    Q or q = heat energy

             m = mass

             C = specific heat  =4.186J/g°C

ΔT = (28°C - 25°C) = 3°C

Now, putting the given values into the above formula as follows.

Q = mCΔT

= 127 × 4.186 × 3

= 1594.86 J or 1.59 kJ    

Therefore, we can conclude that the amount of heat gained by the water is 1.59 kJ    

5 0
2 years ago
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