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3 years ago

Do metals dissolve in water

Chemistry

2 answers:

Oksanka [162]3 years ago

4 0

I think they dont

Explanation:

so I guess metals don't dissolve in water

Nady [450]3 years ago

3 0

yes

include sodium and potassium, dissolve instantly and dramatically in plain water — no stronger acid is needed. The metals react violently with water, releasing and igniting hydrogen gas and causing an explosion.

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If a weak acid is 25 deprotonated at ph 4 what would the pka be

elena-s [515]

4.48

pH=pKa+log([A-/HA])

25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.

4 = pKa + log \frac{.25}{.75}

4 - log \frac{.25}{.75} = pKa

4.48=pKa

6 0

3 years ago

Use the de Broglie's Wave Equation to find the wavelength of an electron moving at 7.3 × 106 m/s. Please show your work. Note: h

ad-work [718]

Answer:

$\lambda=9.96\times 10^{-11}\ m$

Explanation:

The expression for the deBroglie wavelength is:

$\lambda=\frac {h}{m\times v}$

Where,

$\lambda$ is the deBroglie wavelength

h is Planck's constant having value $6.62607\times 10^{-34}\ Js$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

v is the speed of electron.

Given that v = $7.3\times 10^6\ m/s$

Applying in the equation as:

$\lambda=\frac {h}{m\times v}$

$\lambda=\frac{6.62607\times 10^{-34}}{9.11\times 10^{-31}\times 7.3\times 10^6}\ m$

$\lambda=\frac{10^{-34}\times \:6.626}{10^{-25}\times \:66.503}\ m$

$\lambda=\frac{6.626}{10^9\times \:66.503}\ m$

$\lambda=9.96\times 10^{-11}\ m$

6 0

3 years ago

Write the reaction when solid lead(ii) nitrate is put into water:

Whitepunk [10]

Solid lead nitrate in water gives lead oxide and nitric acid

Pb(NO3)2 + H2O ---> PbO + 2 HNO3

7 0

3 years ago

A weather balloon is filled with helium that occupies a volume of 5.00 × 10^4 L at 0.995 atm and 32.0°C. After it is released, i

DaniilM [7]

Answer:

The new volume is 5.913*10^4 L

Explanation:

Step 1: Write out the formula to be used:

Using general gas equation;

P1V1 / T1 =P2V2 /T2

V2 = P1V1T2 / P2T1

Step 2: write out the values given and convert to standard unit's where necessary

P1 = 0.995atm

P2 0.720atm

V1 = 5*10^4 L

T1 = 32°C = 32+ 273 = 305K

T2 = -12°C = -12 + 273 = 261K

Step 3: Equate your values and do the calculation:

V2 = 0.995 * 5*10^4 * 261 / 0.720 * 305

V2 = 1298.475 * 10^4 / 219.6

V2 = 5.913 * 10^4 L

So the new volume of the balloon is 5.913*10^4 L

3 0

4 years ago

How many grams of sodium ions are in na2so4?

LiRa [457]

Answer:

142.04

Explanation:

4 0

3 years ago