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3 years ago

What form does carbon take inside a tree?

Chemistry

2 answers:

daser333 [38]3 years ago

4 0

Carbon dioxide i believe is the answer

miskamm [114]3 years ago

3 0

That would be carbon dioxide

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If the total population of Asian women was 5000in the 2008 how many females suffered a stroke?

Finger [1]

Answer:

Hi there!

I believe you are missing an attachment to this question however

I strongly believe that the answer you are looking for is 85.

Explanation:

If you provide the graph, the rate of strokes in Asian women will be 17 per 1,000 women so all you have to do is multiply 17 by 5 and you get 85

4 0

3 years ago

Which of the following groups of atoms are most likely to have similar properties?

Nezavi [6.7K]

The answer is C.lithium (3), boron (5), and fluorine (9)

6 0

3 years ago

Read 2 more answers

Sugar C12 H22 011 can be classified as a element, heterogeneous mixure, solution, compound

Fantom [35]

Answer:

The answer is a compound.

it is a compound as it contains different elements

Explanation:

5 0

3 years ago

Can someone solve this problem 5

Westkost [7]

Answer:

Step-by-step explanation:

A. Moles before mixing

Beaker I:

Moles of H⁺ = 0.100 L × 0.03 mol/1 L

= 3 × 10⁻³ mol

Beaker II:

Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.

H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]

[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]

[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)

[OH⁻] = 0.01 mol·L⁻¹

Moles of OH⁻ = 0.100 L × 0.01 mol/1 L

= 1 × 10⁻³ mol

B. Moles after mixing

H⁺ + OH⁻ ⟶ H₂O

I/mol: 3 × 10⁻³ 1 × 10⁻³

C/mol: -1 × 10⁻³ -1 × 10⁻³

E/mol: 2 × 10⁻³ 0

You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.

You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.

C. pH

[H⁺] = (2 × 10⁻³ mol)/(0.200 L)

= 1 × 10⁻² mol·L⁻¹

pH = -log[H⁺ ]

= -log(1 × 10⁻²)

= 2

6 0

3 years ago

Calculate the solubility (in g/L) of CaSO 4 ( s ) CaSO4(s) in 0.400 M Na 2 SO 4 ( aq ) at 25 ° C 0.400 M Na2SO4(aq) at 25°C. The

Allushta [10]

Explanation:

Ionization equation for $CaSO_{4}$ is as follows.

$CaSO_{4} \rightarrow Ca^{2+} + SO^{2-}_{4}$

s s s

Now, the expression for the solubility product is as follows.

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

= $s \times s$

= $s^{2}$

As the concentration of $Na_{2}SO_{4}$ is given as 0.4 M.

So, $[Na_{2}SO_{4}] = [SO^{2-}_{4}]$ = 0.4 M

Putting the given values as follows.

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.4$

$[Ca^{2+}]$ = 12.325 \times 10^{-5}[/tex]

Hence, the solubility of $CaSO_{4}$ in $Na_{2}SO_{4}$ is $12.325 \times 10^{-5}$ .

Therefore, solubility of $CaSO_{4}$ in g/ml as follows.

$12.325 \times 10^{-5} \times 136 g/mol$

= 0.0167 g/L

Thus, we can conclude that solubility of $CaSO_{4}$ is 0.0167 g/L.

4 0

3 years ago