statement to analyze and classify as science protoscience pseudoscience or non-science the statement i classified is

One of the hazards of nuclear explosions is the generation of 90Sr andits subsequent incorporation in place of calcium in bones.

Illusion [34]

Explanation:

Initial mass of the isotope 90-Sr in baby= 1.00 μg

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

$t_{\frac{1}{2}}$ = half life of the isotope = 28.1 years

$\lambda$ = rate constant

$N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}$

a) Time taken by the sample, t = 18 years

$N=1.00 \mu g\times e^{-\frac{0.693}{28.1 years}\times 18 years}$

$N = 0.6415 \mu g$

0.6415 μg will remain after 18 years.

b) Time taken by the sample, t = 70 years

$N=1.00 \mu g\times e^{-\frac{0.693}{28.1 years}\times 70 years}$

$N = 0.1779\mu g$

0.1779 μg will remain after 70 years.

3 0

3 years ago

ASAPPP!!!

WITCHER [35]

Answer:

3.16 liters O2

Explanation:

I'll assume the C3H3 is really supposed to read C3H8, for propane.

C3H3(g) + 5 O2 (g) + 3 CO2 (g)+ 4 H20 (g)

C3H8(g) + 5 O2 (g) + 3 CO2 (g)+ 4 H20 (g)

It is unlikely that all the gases involved in this reaction will remain at STP due to the large amount of energy released, we'll assume STP since:

1) We are told STP and no other information is given, and

2) STP makes the problem much easier, since all gasses occupy 22.4 liters per mole of that gas.

The balanced equation tells us we need 5 moles of O2 for every 1 mole of C3H3: a molar ratio of 5/1 (moles O2/moles C3H8).

Calculate moles C3H8 in 0.632 liters of the stuff [metric term for compound]:

(0.632 liters)/(22.4 liters/mole) = 0.0282 moles C3H8

This means we'll need 5 times that number for the amount of O2 required. That comes to 0.1411 moles O2.

Convert that to volume: (0.1411 moles O2)*(22.4 liter/mole) = 3.16 liters O2

6 0

2 years ago

How many grams of calcium chloride will be produced when 28.0 g 28.0 g of calcium carbonate is combined with 14.0 g 14.0 g of hy

zvonat [6]

Answer:

21.3 g of CaCl₂ are produced in this reaction

Explanation:

Reaction is this:

CaCO₃ + 2HCl → CaCl₂ + H₂CO₃

Molar mass salt : 100.08 g/m

Molar mass acid: 36.45 g/m

We have 28 g / 100.08 g/m = 0.279 moles of CaCO₃

We have 14 g / 36.45 g/m = 0.384 moles of HCl

Ratio is 1:2, so 0.279 moles of salt will need the double of moles of acid.

0.279 . 2 = 0.558 moles of acid needed. (I have only 0.384 moles, so the acid is the limiting reactant)

0.384 moles of HCl produce the half of moles of CaCl₂ (This ratio is 1:2) --> 0.192 moles

Molar mass of CaCl₂ = 110.98 g/m

Mol . molar mass = mass → 0.192 m . 110.98 g/m = 21.3 g

5 0

3 years ago

Read 2 more answers

if 25 g of potassium nitrate hydroxide reacts with 25bg of magneisum nitartae to produce potassium nitrate and magneisum hydroxi

Natasha_Volkova [10]

Answer:

0.336 moles of KNO₃ are produced

The limiting reactant is Mg(NO₃)₂

Explanation:

We need to define the reaction to complete this question:

Reactants are: KOH and Mg(NO₃)₂

Products are: KNO₃ and Mg(OH)₂

The reaction is:

2KOH + Mg(NO₃)₂ → 2KNO₃ + Mg(OH)₂

We convert the mass of the reactants, to moles:

25 g / 56.1 g/mol = 0.446 moles

25 g / 148.3 g/mol = 0.168 moles

2 moles of hydroxide need 1 mol of nitrate to react

Then, 0.446 moles of KOH must need (0.446 .1) / 2 = 0.223 moles of nitrate

We do not have enough nitrate, so the Mg(NO₃)₂ is the limiting reagent.

Let's work with the equation and the ratio, which is 1:2

1 mol of Mg(NO₃)₂ produces 2 moles of KNO₃

Then, 0.168 moles of Mg(NO₃)₂ will produce (0.168 . 2) / 1 = 0.336 moles of KNO₃ are produced

5 0

3 years ago

What factor is used to classify acids as strong or weak?

hoa [83]

The extent of dissociation into hydrogen ions in water. A strong acid dissociates completely into hydrogen ions in water whereas a weak acid only dissociates partially

3 0

4 years ago