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WINSTONCH [101]
4 years ago
10

Blood cells. What two effects will such a condition have on the patient?

Chemistry
1 answer:
Rasek [7]4 years ago
6 0
The patient will have less hemoglobin in the blood

The patient will have a reduced oxygen supply to the cells
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A sample of hydrogen gas has an initial pressure of 2.86 atm and an initial volume of 8472 mL If the volume of the
Oliga [24]
<h3>Answer:</h3>

Gas law : Boyle's law

New pressure: 66.24 atm

<h3>Explanation:</h3>

Concept tested: Gas laws (Boyle's law)

<u>We are given,</u>

  • Initial pressure, P₁ = 2.86 atm
  • Initial volume, V₁ = 8472 mL
  • New volume, V₂ IS 365.8 mL

We need to determine the new pressure, P₂

  • According to Boyle's law , the volume of a fixed mass of a gas and the pressure are inversely proportional at constant temperature.
  • That is, P\alpha \frac{1}{V}
  • This means , PV = k (constant)
  • Therefore; P₁V₁ = P₂V₂
  • Rearranging the formula, we can get the new pressure, P₂

P₂ = P₁V₁ ÷ V₂

   = (2.86 atm × 8472 mL) ÷ 365.8 mL

   = 66.24 atm

Therefore, the new pressure is 66.24 atm

5 0
3 years ago
does anyone have any advice for breaking up with your gf because you both lost feelings and after months you have been trying to
inn [45]

Answer:

advise for to break up with her

Explanation:

4 0
3 years ago
Which of the following elements could replace copper in a single replacement reaction?
Likurg_2 [28]
I’m thinking it’s gold because lead isn’t with cooper meaning that if u switch lead with cooper it won’t work at all .
7 0
3 years ago
II. Ionic Equations
mario62 [17]

Answer:

Complete ionic: \begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Net ionic: \begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

Explanation:

Start by identifying species that exist as ions. In general, such species include:

  • Soluble salts.
  • Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2. These three salts will exist as ions:

  • Each \rm AgNO_3\, (aq) formula unit will exist as one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion.
  • Each \rm CaCl_2 formula unit will exist as one \rm Ca^{2+} ion and two \rm Cl^{-} ions (note the subscript in the formula \rm CaCl_2\!.)
  • Each \rm Ca(NO_3)_2 formula unit will exist as one \rm Ca^{2+} and two \rm {NO_3}^{-} ions.

On the other hand, \rm AgCl is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2 (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each \rm AgNO_3\, (aq) formula unit will exist as only one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion. However, because the coefficient of \rm AgNO_3\, (aq)\! in the original equation is two, \!\rm AgNO_3\, (aq) alone should correspond to two \rm Ag^{+}\! ions and two \rm {NO_3}^{-}\! ions.

Do not rewrite the salt \rm AgCl because it is insoluble.

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of \rm Ca^{2+} and two units of \rm {NO_3}^{-}. Doing so will give:

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}.

Simplify the coefficients:

\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

7 0
3 years ago
What volume of a 0.295 M nitric acid solution is required to neutralize 24.8 mL of a 0.107 M potassium hydroxide solution
KengaRu [80]

Answer: 9.00 mL

Explanation:

M_{A}V_{A}=M_{B}V_{B}\\(0.295)V_{A}=(0.107)(24.8)\\V_{A}=\frac{(0.107)(24.8)}{(0.295)} \approx \boxed{9.00 \text{ mL}}

4 0
2 years ago
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