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4 years ago

How many protons does an atom of gold contain?

Chemistry

2 answers:

SpyIntel [72]4 years ago

8 0

There are 79 protons in the nucleus of one gold atom.

snow_tiger [21]4 years ago

4 0

Answer:

79 protons

Explanation:

It is classified as a transition metal. Gold atoms have 79 electrons and 79 protons with 118 neutrons in the most abundant isotope. Under standard conditions gold is a shiny yellow metal. It is very dense and heavy, but also fairly soft.

Hope this helps

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Calculate the mass of aluminum bromide that can be produced from 4.4 mil of bromide

Aloiza [94]

Answer:

197 grams

Explanation:

Without taking into account the number of sig figs you have for your values, you can say that the answer will once again be

197 g of aluminium bromide.

(I hope this is right)

7 0

3 years ago

Do all comets have tails? Explain

Bad White [126]

Answer:

Some (most) has two tails.You can see the tails when the comets go near the sun.There are two types of comet tails,dust and gas ion. Comet tails can be over 600,000 miles (1 million km) long.Because the comet dust particles are so small, they are pushed away from the Sun into a long tail.They are made of ices, such as water, carbon dioxide, ammonia and methane, mixed with dust.There are so much more to discover about comets (which I might not know).

Hope that was helpful.Thank you!!!

5 0

3 years ago

Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets:

Tcecarenko [31]

Answer:

$1.25~mol~H_2O$ and $0.627~mol~N_2$

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ( $N_2H_4$ ) and oxygen ( $O_2$ ). So, we can start with the reaction between these compounds:

$N_2H_4~+~O_2~->~N_2~+~H_2O$

Now we can balance the reaction:

$N_2H_4~+~O_2~->~N_2~+~2H_2O$

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for $N_2H_4$ and 31.99 g/mol for $O_2$ ):

$20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4$

$20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2$

In the balanced reaction we have 1 mol for each reagent (the numbers in front of $O_2$ and $N_2H_4$ are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is $N_2H_4$ .

With this in mind, we can calculate the number of moles for each product. In the case of $N_2$ we have a 1:1 molar ratio (1 mol of $N_2$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2$

We can follow the same logic for the other compound. In the case of $H_2O$ we have a 1:2 molar ratio (2 mol of $H_2O$ is produced by 1 mol of $N_2H_4$ ), so: