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Valentin [98]
3 years ago
8

Compare and Contrast the four types of radiation

Chemistry
1 answer:
Dvinal [7]3 years ago
8 0
The radiation one typically encounters is one of four types: alpha radiation<span>, </span>beta radiation<span>, </span>gamma radiation<span>, and x radiation. </span>Neutron radiation<span> is also encountered in nuclear power plants and high-altitude flight and emitted from some industrial radioactive sources.
.
i dont know if this helps but i hope it does.</span>
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A 9.86 g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid. If 27.5 mL of 0.337 M potassiu
Aneli [31]

Answer:

88.3% of hydroiodic acid is found

7 0
3 years ago
Someone please answer this...
kupik [55]

Answer:

5446.8 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 50 g

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Specific heat capacity (C) = 0.89 J/gºC

Heat (Q) required =?

Next, we shall determine the change in the temperature. This can be obtained as follow:

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 192.4 – 70

ΔT = 122.4 °C

Finally, we shall determine the heat required to heat up the block of aluminum as follow:

Mass (M) = 50 g

Specific heat capacity (C) = 0.89 J/gºC

Change in temperature (ΔT) = 122.4 °C

Heat (Q) required =?

Q = MCΔT

Q = 50 × 0.89 × 122.4

Q = 5446.8 J

Thus, the heat required to heat up the block of aluminum is 5446.8 J

5 0
3 years ago
H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ
Arisa [49]

Answer: help please !!

Explanation:

4 0
3 years ago
Read 2 more answers
3. Fill in: Name the organelle or organelles that perform each of the following functions.
Mekhanik [1.2K]

A. Chloroplasts

B. The cell wall and the vacuole

C. Vacuoles

D. The mitochondrion

8 0
3 years ago
The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is
Feliz [49]
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
5 0
3 years ago
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