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

What is the pH of a solution with an H3O+ concentration of 5.67 x 10-4 M?

Valentin [98]

Answer:

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

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:

Σ $P_g_a_s$ $= P_1+P_2+P_3+...+P_n$

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.

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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:

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

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

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

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

<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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