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Soloha48 [4]
3 years ago
14

A sample of an unknown metal has a mass of 6.557 g. The metal was carefully added to a graduated cylinder containing 10.50 mL of

water. The water level in the graduated cylinder rose to 11.16 mL. What is the density of the unknown metal?​
Chemistry
1 answer:
Levart [38]3 years ago
8 0

Answer:

Explanation:

Be careful. The tricky part of the problem is that there are 4 places of sig digs.

m = 6.557 grams

V = 11.16 - 10.50 = 0.66

density = mass/ volume

density = 6.557/0.66 = 9.935 g/mL

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How many moles of gas are in a 1.0 liter canister if the temperature of the canister is 100 K and the pressure is 100 atmosphere
ladessa [460]

Answer:

n = 12.18 moles

Explanation:

Given that,

The volume of a canister, V = 1 L

The temperature of the canister, T = 100 K

Pressure, P = 100 atm

We need to find the number of moles of gas. Let there are n number of moles. We know that,

PV = nRT

Where

R is gas constant, R = 0.0821 L*atm/mol*K

n=\dfrac{PV}{RT}\\\\n=\dfrac{100\times 1}{0.0821 \times 100}\\\\n=12.18\ moles

Hence, there are 12.18 moles of gas.

4 0
2 years ago
Why does potassium explode when exposed to water?
soldi70 [24.7K]

Answer:

The highly unstable pure sodium or potassium wants to lose an electron and this splits the water atom, producing a negatively charged hydroxide ion and hydrogen and forming an explosive gas that ignites.

Explanation:

4 0
2 years ago
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
Brut [27]

<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

             0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)

             \frac{0.12}{0.0514}=-\ln (0.124 \gamma)

             -2.3346=\ln (0.124 \gamma)

             e^{-2.3346}=0.124 \gamma

             0.0968=0.124 \gamma

             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

6 0
3 years ago
Please help :) thanks
Gnesinka [82]

Explanation:

it is the one you have selected because it is the only solid one

7 0
2 years ago
What is the created by this tilt and the Earth's orbital motion around the Sun?
erica [24]

Answer:

Periodic but not oscillatory

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3 0
2 years ago
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