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makkiz [27]
3 years ago
6

Uranium-235 and uranium-238 are considered isotopes of one another

Chemistry
1 answer:
Bogdan [553]3 years ago
6 0
If you are asking which is the most abundant, Uranium-238 is
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The diagram below represents a sodium atom bonding to a chlorine atom to form sodium chlorine.
Sergio [31]
Sodium Chloride is a compound.
8 0
3 years ago
How much heat is required to raise the<br> temperature of 835g of water from 35.0°C to<br> 65.0°C?
Anna007 [38]

Answer:

25,050 calories.

Explanation:

A calorie is the amount of energy needed to raise the temperature of one gram of water 1 degree centigrade. If we are raising 835 grams of water 30 degrees then we multiply 835*30 to get 25,050 calories.

3 0
3 years ago
What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M = 148.3)?
Butoxors [25]
Mg(NO3)2 ➡️ Mg2+ + 2 NO3-

(32.0g Mg(NO3)2) / (148.3g Mg(NO3)2/mol)* (2 mol NO3- / 1 mol Mg(NO3)2) / (0.425 L) = 1.02 mol/ L NO3-
6 0
3 years ago
What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

7 0
2 years ago
An unknown compound contains only C , H , and O . Combustion of 4.20 g of this compound produced 10.3 g CO 2 and 4.20 g H 2 O .
vodomira [7]

Answer:

C4H8O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 10.3/44 = 0.2341

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.2341= 2.8091

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 4.2/18 = 0.2333

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.2333 = 0.4667 moles

The mass of hydrogen is thus 0.4667* 1 = 0.4667

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 4.2 - 0.4667 - 2.8091 = 0.9242

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.9242/16 = 0.0577625 moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of oxygen

H = 0.4667/0.0577625 = 8

O = 0.0577625/0.0577625 = 1

C = 0.2341/0.0577625 = 4

The empirical formula is thus C4H8O

5 0
2 years ago
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