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stiv31 [10]
3 years ago
13

Consider four beakers labeled a, b, c, and d, each containing an aqueous solution and a solid piece of metal. identity the beake

rs in which a chemical reaction will occur and those in which no reaction will occur.
Chemistry
1 answer:
puteri [66]3 years ago
4 0
To identify in what beakers will a reaction occur, we need to know what aqueous solution is in the four beakers. Then, we also need to know what metals are placed inside those beakers. Now, we can analyze which metals might or will react to the solution in the beakers. 
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If the frequency of an FM wave is 8.85 × 107 hertz, what is the period of the FM wave?
Keith_Richards [23]
NAMASTE

HERE IS YOUR ANSWER:

\boxed{Frequecy = \frac{1}{time \: period} }

Frequency (given) = 8.85× {10}^{7}hz

time period =
\frac{1}{8.85 \times {10}^{7}}

= 0.112 × {10}^{-7}

= 1.12 × {10}^{-8}
7 0
3 years ago
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PLEASE PLEASE PLEASE HELP ME
andrey2020 [161]

3. Cl₂ + 2KI --> 2KCl + I₂

Cl = 2        Cl = 2

K = 2         K = 2

I = 2           I = 2

4. 2NaCl --> 2Na + Cl₂

Na = 2      Na = 2

Cl = 2       Cl = 2

5. 4Na + O₂ --> 2Na₂O

Na = 4      Na = 4

O = 2        O = 2

6. 2Na + 2HCl --> H₂ + 2NaCl

Na = 2         Na = 2

H = 2           H = 2

Cl = 2          Cl = 2

7. 2K + Cl₂ --> 2KCl

K = 2         K = 2

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8 0
3 years ago
How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?
andrezito [222]

Answer:

= 29.64 g  NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

                             = 1.55 M × 0.225 L

                             = 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

   = 0.34875 moles × 84.99 g/mol

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5 0
3 years ago
The volume of a gas held at constant temperature varies indirectly as the pressure of the gas. If the volume of a gas is 1200 cu
kvasek [131]

Answer:

Volume of the gas at 500 mm Hg pressure is 960 cm^{3}

Explanation:

Let's assume the gas behaves ideally.

According to combined gas law for an ideal gas-

                  \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

Where P_{1} and  P_{2} are initial and final pressure of the gas respectively.  V_{1} and  V_{2} are initial and final volume of the gas respectively.  T_{1} and  T_{2} are initial and final temperature of the gas in kelvin respectively.

Here T_{1} = T_{2}, V_{1} = 1200 cm^{3}, P_{1} = 400 mm Hg, P_{2} = 500 mm Hg

So, V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}

Hence V_{2}=\frac{400\times 1200}{500}[tex]cm^{3}=960 cm^{3}[/tex]

Hence volume of the gas at 500 mm Hg pressure is 960 cm^{3}

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Which are the organs of the digestive system
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Explanation:

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3 years ago
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