Ask question

Ask question

All categories

2 years ago

5

Help pleaseeeeeeeeeeeeee!!?

2 answers:

melisa1 [442]2 years ago

4 0

Answer:

There are 146 neutrons in Uranium-238

There are 90 electrons in Thorium-234

vodka [1.7K]2 years ago

4 0

146 N in uranium and 90 E in Thorium

You might be interested in

A sample of 10.5 g of nitrogen reacts with 20.2 g of hydrogen to produce ammonia. (36 points

BartSMP [9]

Answer:

Explanation:

A. 55

B. 43

C. 14

D. 29

6 0

2 years ago

How does distance affect work pls quick its 15 points pls keep it short pls help!!!!

ra1l [238]

The simple equation used to calculate work is force multiplied by distance, thus as this is the case increasing the distance by a certain amount, assuming the force applied to the object is constant, the amount of work you are doing on the box for instance pushing it, is going to be greater

Since you are pushing the box with the same force covering a greater distance with the force.

4 0

3 years ago

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the <u>reaction</u>:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

3 years ago

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is $1.743\times 10^2g$

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

3 years ago

Why are atoms more stable when combined?

shepuryov [24]

Answer:

D

Explanation:

3 0

3 years ago