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2 years ago

In which type of radioactive decay does the nucleus become more stable without changing its identity?(1 point)

Chemistry

2 answers:

san4es73 [151]2 years ago

8 0

Answer:

gamma decay

Explanation:

____ [38]2 years ago

3 0

Answer:

the answer would be the first one, gamma decay

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I am a mental with 28 electrons

musickatia [10]

I am a Nickel.
You're welcome.

3 0

3 years ago

When you have analyzed your data and you form a _____________ ?

MAVERICK [17]

Hypothesis or outline

8 0

3 years ago

Caffeine (C8H10N4O2) is a stimulant found in coffees and teas. When dissolved in water, it can accept a proton from a water mole

Olenka [21]

Answer:

See figure 1

Explanation:

If we want to find the acid and the Brønsted-Lowry base, we must remember the definition for each of these molecules:

-) Acid: hydrogen donor

-) Base: hydrogen acceptor

In the caffeine structure, we have several atoms of nitrogen. These nitrogen atoms have the ability to accept hydronium ions ( $H^+$ ). Therefore the caffeine molecule will be the base since it can accept

If caffeine is the base, the water must be the acid. So, the water in this reaction donated a hydronium ion.

Thus, caffeine is the base and water the acid. (See figure 1)

3 0

3 years ago

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which

Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride = 0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be: $= \dfrac{5 \ kg }{1590 \ kg/m^3}$

= 0.0031 m³

Suppose $\beta$ is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

$\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT$

$In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)$

$V_2 = V_1 \times exp (\beta (T_2-T_1))$

$V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)$

$V_2 = 0.003175 \ m^3$

However; the workdone = -PdV

$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

The heat energy Q = Δ h

$Q = mC_p(T_2-T_1)$

$Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20$

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0

3 years ago

A student pours 44.3 g of water at 10Â°C into a beaker containing 115.2 g of water at 10Â°C. What are the final mass, temperatur

Agata [3.3K]

Answer:

Final mass = 159.5 g

Final temperature = 10 C

Final density = 1.00 g/ml

Explanation:

Given:

Beaker 1:

Mass of water = 44.3 g

Temperature = 10 C

Beaker 2:

Mass of water = 115.2 g

Temperature = 10 C

Density of water at 10C = 1.00 g/ml

To determine:

The final mass, temperature and density of water

Calculation:

$Final\ mass\ of \ water = Beaker\ 1 + Beaker\ 2 = 44.3 + 115.2 = 159.5 g$

Since there is no change in temperature, the final temperature will be 10 C

Density of a substance is an intensive property i.e. it is independent of the mass. Hence the density of water will remain constant i.e. 1.00 g/ml

3 0

3 years ago