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Triss [41]
3 years ago
12

.

Chemistry
1 answer:
Mila [183]3 years ago
7 0

Answer:

Explanation:

Half life of Nobelium-253 is 97 seconds . That means after every 97 seconds half of the Nobelium amount will be disintegrated .

Time taken in bringing the sample to laboratory = 291 seconds

291 second = 291 / 97 half life

n = 3

N = N_0 (\frac{1}{2})^n

N₀ is original mass , N is mass after n number of half life.

N = 5 mg x (\frac{1}{2})^3

= .625 mg

Only  0.625 mg of Nobelium-253 will be left .

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13. Describe briefly how you figured out what the name of the element is:
zloy xaker [14]
Answer:
13) between the nucleus, there is a proton and electron number given, (looking at the periodic table, it gives you the name of the element)

14) an ion, in any element has a POSITIVE charge no matter what.. the numbers depend on what it element the ion is with..
7 0
3 years ago
How many kilocalories of heat are released when 75 g of steam at 100 degrees Celsius is converted to ice at 0 degrees Celsius?
VikaD [51]

Answer:

∴ Q = -7.52kCal

Explanation:

Using the formula for specific heat capacity:

Q = mcΔT

where ΔT = change in temperature (final - initial)  = (0 - 100)°C = -100°C

m = mass (g) = 75g

c = specific heat capacity = 4.2 J/g°C in water

⇒ Q = 75 × 4.2 × -100

= -31,500J

But 1J - 0.000239kCal

<u>∴ Q = -7.52kCal</u>

<u />

Let me know if I can be of further assistance.

5 0
3 years ago
Patient added a 17-g measured dose of polyethylene glycol 3350 (MIRALAX) to 180 mL of water to use as a laxative. If the volume
Sergio [31]

Answer: Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

Specific gravity of mixture = 1.01

Explanation:

The specific gravity of an object or substance is the ratio of the density of that object or substance to that of a reference substance usually water.

<em>Specific gravity = density of substance/density of water</em>

The density of a substance is the ratio of the mass of that substance to is volume.

<em>Density = Mass/volume</em>

Density of water 1.00 g/mL,  

Apparent density of ethylene glycol 3350 = mass/volume

volume of ethylene glycol = volume of mixture - volume of water

volume of ethylene glycol = 195.6 - 180) = 15.6mL

Apparent density of ethylene glycol 3350 = 17g/15.6mL= 1.09g/mL

<em>mass of water =  density of water * volume of water</em>

mass of water = 1.00g/mL * 180mL = 180 g

Mass of solution = mass of substance + mass of water = (17 + 180)g = 197g

Density of mixture = mass of mixture / volume of mixture

Density of mixture = 197 g / 195.6 mL = 1.01 g/mL

specific gravity of mixture = density of mixture / density of water

specific gravity of mixture = 1.01g/mL / 1.00g/mL

specific gravity of mixture = 1.01

8 0
3 years ago
Help please ill do anything really!!!!!!!
Volgvan

Answer:

The first one is air the second is decreases the third is water the fourth is gas and the last is liquid.

Explanation:

Hope it helps.

5 0
3 years ago
The concentration of hydrogen peroxide solution can be determined by
max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0
3 years ago
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