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Yuri [45]
3 years ago
10

Convert 50 gr of water into moles and into molecules of H2O.

Chemistry
1 answer:
Alina [70]3 years ago
5 0
Uh I can help you with the converting into moles, 50 grams= 2.775421, sorry if I didn’t help much :/
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A sample contains 16 mg of polonium-218. After 12 minutes, the sample will contain 1.0 mg of polonium-218. What is the half life
mariarad [96]

Answer:

Half-life = 3 minutes

Explanation:

Using the radioactive decay equation we can solve for reaction constant, k. And by using:

K = ln2 / Half-life

We can find half-life of polonium-218

Radioactive decay:

Ln[A] = -kt + ln [A]₀

Where:

[A] could be taken as mass of polonium after t time: 1.0mg

k is Reaction constant, our incognite

t are 12 min

[A]₀ initial amount of polonium-218: 16mg

Ln[A] = -kt + ln [A]₀

Ln[1.0mg] = -k*12min + ln [16mg]

-2.7726 = - k*12min

k = 0.231min⁻¹

Half-life = ln 2 / 0.231min⁻¹

<h3>Half-life = 3 minutes</h3>

5 0
3 years ago
A salt wedge forms in an estuary because
kherson [118]

Answer:

a rapidly flowing river discharges into the ocean where tidal currents are weak.

Explanation:

The force of the river pushing fresh water out to sea rather than tidal currents transporting seawater upstream determines the water circulation in these estuaries.

6 0
3 years ago
Which equation represents the combined gas law?
stealth61 [152]

Answer:

P1V1/T1= P2V2/T2

Explanation:

Combined gas law involves Boyle's law and Charles law altogether with the formula of Boyle's law as P1V1=P2V2

formula for charles law as V1/T1=V2/T2

so when combined form P1V1/T1=P2V2/T2

4 0
3 years ago
what is the percent by mass of NaHCO3 in a solution containing 20.0 g of NaHCO3 dissolved in 600.0 ml of water
Ne4ueva [31]
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%

m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water

%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
7 0
3 years ago
A 0.590 gram sample of a metal, M, reacts completely with sulfuric acid according to:M(s) +H2SO4(aq) --&gt; MSO4(aq) +H2(g)A vol
photoshop1234 [79]

Answer:

MM = 58.41 g

Explanation:

First, the data we have is according to the hydrogen which is exerting pressure. To solve this, we need to use the ideal gas equation:

PV = nRT (1)

the molar mass of any compound is calculated like this:

MM = m/n (2)

So, from (1) we solve for the moles (n) and then, this value is replace in (2).

However, before we do all that, we need to gather all the correct data.

All the species in the reaction are solid or aqueous state, with the exception of hydrogen, which is gaseous. Hydrogen is collected over water, therefore, is exerting some pressure too. The problem is not indicating if the acid or any other species is exerting pressure, so we will assume that only hydrogen and water are exerting pressure.

The total pressure exerted by the system would be:

P = Pw + PH2 (3)

We already know the total pressure which is 756 torr.

This experiment is taking place at 25 °C (298.15 K), and at this temperature, we have a reported value for water pressure which is 23.8 Torr.

Let's solve for PH2:

PH2 = P - Pw

PH2 = 756 - 23 = 733 Torr

Now, with this value, and the volume and temperature, we can calculate the moles of H2:

n = PV/RT

But first, let's convert the pressure to atm:

PH2 = 733 Torr / 760 torr * 1 atm = 0.9644 atm

now, solving for n:

n = 0.9644 * (0.255) / 0.082 * 298.15

n = 0.0101 moles

Now that we have the moles, we know that the metal and the hydrogen has a mole ratio of 1:1 according to the reaction, so, this means that:

moles M = moles H2 = 0.0101 moles

We have the moles of the metal and the mass, we can calculate the molar mass using expression (2):

MM = 0.590/0.0101

MM = 58.41 g/mol

This is the molar mass of the metal

8 0
3 years ago
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