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bezimeni [28]
3 years ago
6

Which of the following radiation sources provides the greatest radiation dose?

Chemistry
1 answer:
Sedbober [7]3 years ago
6 0
I believe the answer is background radiatin
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Bradley was working in the lab. He accidentally poured a container of concentrated hydrochloric acid into his container of disti
nexus9112 [7]

Answer is: The solution has now become a good conductor of electricity.

Hydrochloric acid (HCl) dissociate on positive ions or cations of hydogen (H⁺) and negative ions or anions of chlorine (Cl⁻) accordinf to balanced chemical reaction:

HCl(aq) → H⁺(aq) + Cl⁻(aq).

When there are free cations and ions, water solution can conduct electricity.

4 0
3 years ago
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What are the two categories of chemical sedimentary rock?
AnnyKZ [126]
Calcite and calcite?
4 0
3 years ago
What is the temperature -71 oC expressed in Kelvin?<br> Group of answer choices
satela [25.4K]

Answer: 202.15

Explanation:

-71°C + 273.15

= 202.15K

5 0
4 years ago
How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 163 mL of coffee at 97.9°C so that the r
Stolb23 [73]

Answer:

248 mL

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.

Qw + Qc = 0

Qw = -Qc [1]

We can calculate each heat using the following expression.

Q = c × m × ΔT

where,

  • c: specific heat
  • m: mass
  • ΔT: change in the temperature

163 mL of coffee with a density of 0.997 g/mL have a mass of:

163 mL × 0.997 g/mL = 163 g

From [1]

Qw = -Qc

cw × mw × ΔTw = -cc × mc × ΔTc

mw × ΔTw = -mc × ΔTc

mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)

mw × 29.0°C = 163 g × 43.9°C

mw = 247 g

The volume corresponding to 247 g of water is:

247 g × (1 mL/0.997 g) = 248 mL

8 0
3 years ago
Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
Cloud [144]

Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

Solution : Given,

Equilibrium constant, K_c=1.8\times 10^{-5}

Initial concentration of HC_2H_3O_2 = 0.260 m

Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

                  HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)

Initially                0.260 m                       0                 0

At equilibrium    (0.260 - x)                     x                 x

The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

Now put all the given values in this expression, we get

1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}

By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

4 0
3 years ago
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