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wlad13 [49]
3 years ago
11

What are two things that happened as a blood passes the lungs?

Chemistry
1 answer:
amm18123 years ago
7 0

Answer:

blood travels to the pulmonary arteries

Explanation:

This is called the pulmonary circulation. From your pulmonic valve, blood travels to the pulmonary arteries and eventually to tiny capillary vessels in the lungs.

You might be interested in
The parent element 247 95 am decays by beta decay which daughter element will formn
ziro4ka [17]
Atomic mass of the parent element =247,
atomic number of the parent element = 95

In the process of β-decay electron leaves the nucleus, so instead of one neutron we get one proton.

Mass of proton≈mass of neutron,
so atomic mass will not change.

Charge of proton =+1, and charge of neutron = 0.
So, we will get atomic number increased by one.

New element (daughter) will have 
atomic mass = 247, 
and atomic number= 95+1=96

Number 95 - Am (parent),
number 96 - Cm(daughter),

So, from Am-247 we will get Cm-247.


6 0
3 years ago
Which has a larger molar mass, salt (NaCl) or sugar (C12H22011) – how can you tell?
uysha [10]

Answer:

Explanation:

Molar mass of NaCl = (23+35.5)

Molar mass of NaCl = 58.5g/mol

Molar mass of C12H22011

= 12(12) + 22(1) + 16(11)

= 144 +22 + 176

= 342g/mol

4 0
3 years ago
Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume
pishuonlain [190]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

             mass = 0.508 g,               Volume = 0.175 L

             Temperature = (25 + 273) K = 298 K,       P = 1 atm

As per the ideal gas law, PV = nRT.

where,  n = no. of moles = \frac{mass}{\text{molar mass}}

Hence, putting all the given values into the ideal gas equation as follows.

               PV = \frac{mass}{\text{molar mass}} \times RT            

           1 atm \times 0.175 L = \frac{0.508 g}{\text{molar mass}} \times 0.0821 L atm/ K mol \times 298 K  

                            = 71.02 g

As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of Cl_{2} is 70.8 g/mol or 71 g/mol (approx).

Thus, we can conclude that the gas is most likely chlorine.

4 0
3 years ago
Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.
Rudiy27
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

                                ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where: 
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg


For the problem, 
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius
5 0
4 years ago
Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

  • \text{NH}_4 \text{Cl} \; (s): -2 \; a = 1
  • \text{NH}_3\; (g): -2 \; b = -1
  • \text{HCl} \; (g): 2 \; c = -1

Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

3 0
3 years ago
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