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3 years ago

Identify each type of titration curve. note that the analyte is stated first, followed by the titrant.

1 answer:

8 0

Each of the type of titration curve are as follows: Strong acid-strong base= starts small than increases<span>Weak acid-strong base= low then high (small difference) Weak base- strong acid= high to low. Have in mind that titraton curves generally contain the volume of the titrant as the independent variable and the ph of the solution as the dependent variable. </span>

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Which group of elements has the most similar chemical properties? (1 point)

enot [183]

Answer:

carbon nitrogen and oxygen

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3 years ago

What volume of 0.200 M H2SO* will be needed to neutralize<br><br> 50.00 mL of 0.178 MNaOH?

mixas84 [53]

Answer:

10.00 mL

Explanation:

Please, do not forget to write the dot and 2 of Zeros after the number 10 in the answer!

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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4 years ago

Select the statements that correctly describe an asteroid. Check all that apply

stich3 [128]

Answer:

Explanation:

Most of this ancient space rubble can be found orbiting the Sun between Mars and Jupiter within the main asteroid belt. Asteroids range in size from Vesta – the largest at about 329 miles (530 kilometers) in diameter – to bodies that are less than 33 feet (10 meters) across.

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3 years ago

A female spinach plant with flat (Ff) leaves is crossed with (pollination and fertilization occur) a male spinach plant with cri

Vedmedyk [2.9K]

Answer:

Question

A female spinach plant with flat (Ff) leaves is crossed with (pollination and fertilization occur) a male spinach plant with crinkly (ff) leaves. Explain wha

Explanation:

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3 years ago

The heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its temp

LekaFEV [45]

Answer:

$Q=9.2x10^5J$

$t=614s=10.2min$

Explanation:

Hello,

In this case, we can compute the energy by using the following formula for air:

$Q=nCp\Delta T$

Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:

$n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol$

Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:

$Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J$

Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:

$t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min$

Regards.

8 0

3 years ago