Feature E on the map is composed of glacial till.What type of glacial till is E, and how did it form?

Suppose that 100 grams of water at 50.0°C is placed in contact with 200 grams of iron at 30.0°C. The final

wel

Answer:

The answer would be 1.5 kJ.

Explanation:

When you use the equation q = m x c x ∆T you will be able to find the energy gained or lost. The data for the water in this case is just there to distract you so ignore it. :D

4 0

2 years ago

Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s

Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, $2.67\times 10^2mol/L$

Explanation :

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = molar solubility of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 0.2 atm = 1.97×10⁻⁶ Pa

$k_H$ = Henry's law constant = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

$C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)$

$C_{O_2}=8.55\times 10^3g/L$

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = $\frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L$

Therefore, the molar concentration of oxygen is, $2.67\times 10^2mol/L$

8 0

3 years ago

How many moles of H 2 can be formed if a 3.24 g sample of M g reacts with excess H C l ?

Ganezh [65]

Answer:

0.135 mole of H2.

Explanation:

We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:

Mass of Mg = 3.24 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 3.24/24

Mole of Mg = 0.135 mole

Next, we shall write the balanced equation for the reaction. This is illustrated below:

Mg + 2HCl —> MgCl2 + H2

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.

Thus, 0.135 mole of H2 can be obtained from the reaction.

3 0

3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0

2 years ago

If 100cm3 of O2 diffused in 4s and 50cm3 of gas Y diffused in 3s. calculate the relative molecular mass of gas X (O=16).

sertanlavr [38]

Answer:

T2/T=√M1√M2

T2=3 T1=4 M1=O2=32

M2=?

4/2=√M2/√32

=1.3

Explanation:

8 0

2 years ago