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Ray Of Light [21]
3 years ago
6

Uranium is classified as what type of element?

Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

actinide metal

Explanation:

That is what i found when  I seartcht up i

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How would you explain the close similarities between
PolarNik [594]
<span>They have similar or common ancestor. The similarities inherited from the common ancestor as in the case of homologous organs. In evolutionary biology a group of organisms share a common descent if they have a common ancestor. There are many other examples like in the case of hemoglobin to prove the theory of common ancestor</span>
4 0
4 years ago
N2 +3H2 → 2NH3 What volume of ammonia at STP is produced if 30.0 g of nitrogen gas is reacted with an excess of hydrogen gas?
WARRIOR [948]

Answer:

V= 48L

Explanation:

Moles of N2 = m/M = 30/28= 1.07moles

From the equation of reaction

N2 +3H2 → 2NH3

1mole of N2 produces 2mole of NH3

Hence

1.07moles will produce 1.07×2= 2.14 moles

At STP, 1mole occupy 22.4L

Hence volume of N2 produced = 2.14×22.4= 48 L

3 0
4 years ago
4) A 4.00 L balloon is filled with 0.297 moles of helium gas with a pressure
Nadya [2.5K]

Answer: 149

Explanation:

Using the ideal gas equation; PV = nRT

P= Pressure = 0.910 atm,     T= Temperature = ?

V= Volume = 4.0L                  R = Gas constant = 0.08206 L.atm/mol/K

n = number of moles = 0.297

Making 'T' the subject of the formular, we have;

T = P V/ n R   =  0.910  x 4  /  0.297 x 0.08206

                       =  149

6 0
3 years ago
3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent wou
juin [17]
<span>2Al + 3Br2 -------------> 2AlBr3

</span>3 g Al = 0.11 mol Al. 

<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>

<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>

<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g. 
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>


</span>
8 0
4 years ago
10. Copper(i) bromide reacts with magnesium metal: 2 CuBr + Mg → 2 Cu + MgBrz
velikii [3]

Answer:

72.6 grams

Explanation:

I got this answer through stoichiometry.  For every 1 mole of Mg, 2 moles of CuBr are consumed.  Because of this, multiply the moles of Mg by 2.  Then, convert moles to grams.

5 0
3 years ago
Read 2 more answers
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