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2 years ago

What is the name for Cl₇Br₄ *

Chemistry

1 answer:

Margarita [4]2 years ago

4 0

Answer: no se wey jaja que le pusiste tu ?

Explanation:

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Estimate the air pressure at an altitude of 5km

r-ruslan [8.4K]

The answer is 100 kg

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3 years ago

Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Misha Larkins [42]

Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

$moles HCl = moles KOH \\ \\ cHCl*vHCl=cKOH*vKOH \\ \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}= 541,54mL$

So, the required volume of HCl is 541,54 mL

Have a nice day!

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3 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,

1 ton=907.185 kg, thus,

$3.3480\times 10^{4} tons=3.037\times 10^{7} kg$

Thus, production of $CO_{2}$ will be $3.037\times 10^{7} kg/ day$ .

The specific volume of $CO_{2}$ is $0.0120 m^{3}/kg$ .

Volume of $CO_{2}$ produced per day can be calculated as:

$V=Specific volume\times mass$

Putting the values,

$V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day$

Thus, volume of $CO_{2}$ produced per year will be:

$V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year$

Thus, in 4 year volume of $CO_{2}$ produced will be:

$V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}$

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3 years ago

What’s is solar energy

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2 years ago

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What features of this model will help Armando answer the question?

ASHA 777 [7]

Answer:

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3 years ago