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Maru [420]
3 years ago
6

When reptiles and fish young are they look like __ similar to their parents? Please hurry!!

Chemistry
1 answer:
WARRIOR [948]3 years ago
8 0
Fishes yes according to the age, reptiles yes
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What is the pH of a solution with an H3O+ concentration of 5.67 x 10-4 M?
Valentin [98]

Answer:

<h2>3.25 </h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ { H_3O}^{+}]

From the question we have

ph =  -  log(5.67 \times  {10}^{ - 4} )  \\  = 3.2464

We have the final answer as

<h3>3.25 </h3>

Hope this helps you

3 0
3 years ago
How moles are in 45.7g of CH3?
miskamm [114]

first find the atomic weight of CH3 which would be

atomic weight: 12.011 (3×1.008) = 36.32 g/mol

then find the moles in the given mass

36.32 ÷ 45.7 = 0.794

I HOPE I'M NOT WRONG I HAVENT DONE CHEM IN SO LONG

6 0
3 years ago
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
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
For years, surgeons have had great success using _____.
FrozenT [24]
C the artifical heart
6 0
3 years ago
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