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3 years ago

An engineer uses aluminum to build an airplane rather than composite

Chemistry

1 answer:

lesantik [10]3 years ago

5 0

Answer:

Explanation:

I would assume A because it says in the problem statement that chooses aluminum because he is worried that composite will suddenly fail... therefore he would use aluminum because it would bend before breaking and therefore show warning.

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Which diagram best represents a gas in a closed container?

Artist 52 [7]

Answer:

Explanation:

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3 years ago

Major species present when dissolved in water ammonium iodide

Yuri [45]

Ammonium iodide, or NH₄I, is a salt. Since all salts are strong electrolytes, when ammonium iodide is dissolved in water, it would dissociate into ammonium ions and iodide ions. Hence, the major species present would be: <em>NH₄⁺ and I⁻ ions</em>.

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4 years ago

Larger companies often need many workers who do different steps in the transportation process, instead of each member doing all

Ksju [112]

Answer:

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Explanation:

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3 years ago

The half-life of bismuth-210, 210bi, is 5 days. (a) if a sample has a mass of 184 mg, find the amount remaining after 15 days.

ASHA 777 [7]

Answer:- 23.0 mg

Solution:- Radioactive decay obeys first order kinetics and the first order kinetics equation is:

$lnN=-kt+lnN_0$

where, $N_0$ is the initial amount of radioactive substance and N is it's amount after time t. k is the decay constant.

From given information, Original amount, $N_0$ of the radioactive substance is 184 mg and we are asked to calculate the amount N after 15 days. It means, t = 15 days

Half life is given as 5 days. From the half life, we could calculate the decay constant k using the equation:

$k=\frac{0.693}{t_1_/_2}$

where, $t_1_/_2$ is the symbol for half life. let's plug in the value of half like to calculate k:

$k=\frac{0.693}{5days}$

$k=0.1386day^-^1$

Let's plug in the values in the first order kinetics equation and solve it for N:

$lnN=-0.1386day^-^1(15days)+ln184mg$

$lnN=-2.079+5.215$

lnN = 3.136

$N=e^3^.^1^3^6$

N = 23.0 mg

So, 23.0 mg of Bi-210 would be remaining after 15 days.

3 0

4 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

6 0

3 years ago