All categories

2 years ago

Explain how our body systems work together to get oxygen into and around our body.

1 answer:

7 0

Answer: (1) Inhales (breathes in) Oxygen - good for the body - gives it to the Circulatory System to be transported throughout the body through the blood. (1) Digestive System gets nutrients (good) from food and hands it over to the blood and Circulatory System then carries those nutrients where they need to go.

You might be interested in

Mass box A = 10 grams; Mass box B = 5 grams; Mass box C—made of one A and one B Mass ratio A/B =

LUCKY_DIMON [66]

Answer:

mass ratio of A/B is 2:1

Explanation:

Since the mass of box A = 10g

mass of box B = 5g

Mass of box C = mass of box A + mass of box

A ratio compares two quantities. To find the ratio of the two boxes:

Ratio of A to B = $\frac{mass of A}{mass of B}$

Ratio of A to B = $\frac{10grams}{5grams}$ = 2

The mass ratio is 2:1 i.e box A has twice the mass of B

8 0

3 years ago

What's the charge on Ni in the compound NiCi3? *

ss7ja [257]

Answer:

Explanation:

Chlorine is anion with a -1 charge. But they are three chlorine atoms.

-1 * 3 = -3

So they have a -3 charge.

So to balance the compound, the nickel has to be a cation with a +3 charge.

-3 + 3 = 0

Furthermore, a chemical bond always has a 0 charge. Remember that.

Hope it helped! Rate my answer a 5 star if correct.

8 0

3 years ago

Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

78.9 + 890.43 - 21 = 9.5 x 10^2

maria [59]

948 or 9.48 x 10^2

There are two sets of rules for significant figures

• One set for addition and subtraction

• Another set for multiplication and division

You used the set for multiplication and division.

This problem involves addition and subtraction, and the rule is

The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.

Thus, we have

78.9

+890.43

-21.

= 948.33

The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.

To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.

8 0

3 years ago

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

3 years ago