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2 years ago

36cm/s to miles per hour (mph)

Chemistry

1 answer:

Ipatiy [6.2K]2 years ago

7 0

Answer:

0.8 mph

Explanation:

36 cm 1 m 1 km 0.62 mi 60 sec 60 min 0.8 mi

---------- X ------------ X ------------- X -------------- X ------------ X -------------- = -----------

sec 100 cm 1000 m 1 km min 1 Hr Hr

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An object in space is made of gas and gives off light and heat energy. We can only see it at night. Which object is it? (2 point

strojnjashka [21]

Answer:

stars

earth isn't made of gas. (and you can always see it)

moon us not gas, or give off heat

sun is only visible during the day

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2 years ago

What is the volume in ML

BabaBlast [244]

Answer:1 mL

Explanation:

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3 years ago

Which solution is the most concentrated? 2.0 mL of 10 M H2SO4, where H2SO4 has a molar mass of 98 g/mol 5.0 mL of 1.0 M PbSO4, w

marishachu [46]

Answer: the option 4) 2.0 mL of 10.5 M H₂O₂, where H₂O₂ has a molar mass of 34 g/mol.

Its concentration is 10.5 M.

Explanation:

1) The unit M means molar. It is the molarity of the solution.

Molartity is the concentration of the solution expressed as number of moles of solute per liters of solution.

The formula of molarity, M, is:

M = number of moles of solute / volume of solution in liters

2) 2.0 mL of 10 M H₂SO₄, where H₂SO₄ has a molar mass of 98 g/mol

⇒ concentration is 10 M

3) 5.0 mL of 1.0 M PbSO₄, where PbSO₄ has a molar mass of 303 g/mol

⇒ concentration = 1.0 M

4) 2.0 mL of 10.5 M H₂O₂, where H₂O₂ has a molar mass of 34 g/mol

⇒ concentration is 10.5 M

5) 100 mL of 10 M NaCl, where NaCl has a molar mass of 58 g/mol

⇒ concentration is 10 M

7 0

3 years ago

Read 2 more answers

Identify which statement about a solution is correct. Group of answer choices The solution has the same physical state as the ph

zavuch27 [327]

Water can only be the solvent, never the solute is the statement about a solution that is correct.

The solute is the substance that, in general, is found in less quantity and that dissolves in the mixture.

The solvent, on the other hand, is the substance that usually appears in greater quantity and where the solute dissolves.

The most usual thing is that the solvent is the substance that establishes the physical state of the solution.

Water is considered a universal solvent, since it is the liquid that dissolves the most substances, a quality linked to its status as a “polar molecule”, this is due to its ability to form hydrogen bonds with other substances.

Solubility depends on the properties of a solvent that allow it to interact with a solute in a stronger way than the solvent particles do with each other.

In particular, the polar character of water makes it an excellent solvent for polar and ionic solutes, which are called hydrophilic.

Therefore, we can conclude that water can only be the solvent, never the solute is the statement about a solution that is correct.

Learn more here: brainly.com/question/2835889

5 0

2 years ago

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g

Tanya [424]

Answer:

(A) $\Delta H^{\circ }_{r}= -144 kJ$

(B) $\Delta H^{\circ }_{r}= - 2552kJ$

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

$1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)$

$1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)$

Now, multiplying equation (a) with 2:

⇒ $N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)$

Then equation b is reversed and multiplied with 2:

$2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)$

Now by adding the equation (a) and equation (b), we get:

⇒ $2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)$

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

$\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}$

$= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ$

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

$B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)$

$2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)$

$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)$

$H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)$

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

$6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)$

$4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)$

$6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)$

$6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)$

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

$\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }$

$= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ$

4 0

3 years ago