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Oxana [17]
3 years ago
6

If you start with 42 g of Fe-53, how much is left after 8.51 minutes? I​

Chemistry
2 answers:
lidiya [134]3 years ago
5 0

Answer:

2.58 good luck with everything

Paladinen [302]3 years ago
4 0

Answer:

2:38

Explanation:

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In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide
Black_prince [1.1K]

Answer:

Explanation:

The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:

ΣP_g_a_s = P_1+P_2+P_3+...+P_n

The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:

P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr

Solve for Po2:

P_o_2=(760-582.008-0.285) \ torr = 177.707 \ torr

Thus, the partial pressure of diatomic oxygen is 177.707 torr.

<u><em>If you liked this solution, hit Thanks or give a Rating!</em></u>

4 0
3 years ago
Identify the element using the Bohr model below.​
Maurinko [17]
Nitrogen I believe . I need 20 characters.
3 0
3 years ago
What is the net ionic equation for ammonia and phosphoric acid?
olganol [36]
 N<span>et ionic equation for ammonia and phosphoric acid</span>  is 
3 NH4OH + H3PO4 >> (NH4)3PO4 + 3 H2O
hope this helps
7 0
4 years ago
Which of these is NOT true about electromagnets?
Alex17521 [72]

Answer:

Permanent = false

Explanation:

All of the other choices are true

4 0
2 years ago
What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according
ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

<u>Step 1: </u>The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

<u>Step 2: </u>Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

<u>Step 3: </u>Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

<u>Step 4: </u>Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M  = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0
3 years ago
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