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2 years ago

12

Soo Hwan found a flowering plant at school and used a classification key to help him classify the plant. The flowers of the plan

t are not clustered, and they have a deep brown center. What type of plant did Soo Hwan find?

1 answer:

KIM [24]2 years ago

8 0

Answer:

Black eyed susan

Explanation:

...

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Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient

romanna [79]

Answer:

Coefficient of $H^{+}(aq)$ is more than 4

Explanation:

Oxidation: $Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance charge: $Sn^{2+}(aq)-2e^{-}\rightarrow Sn^{4+}(aq)$ ......(1)

Reduction: $Cr_{2}O_{7}^{2-}(aq)\rightarrow Cr^{3+}(aq)$

Balance Cr: $Cr_{2}O_{7}^{2-}(aq)\rightarrow 2Cr^{3+}(aq)$
Balance O and H in acidic medium: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$
Balance charge: $Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)$ .......(2)

$[3\times Equation-(1)]+Equation(2)$ gives balanced equation:

$3Sn^{2+}(aq)+Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)\rightarrow 3Sn^{4+}(aq)+2Cr^{3+}(aq)+7H_{2}O(l)$

So coefficient of $H^{+}(aq)$ is more than 4

3 0

2 years ago

Neutral atoms of all the isotopes of the same element have

Viktor [21]

Answer:

electron = proton

Explanation:

6 0

2 years ago

Read 2 more answers

A quantity of 200 mL of 0.862 M HCl (aq) is mixed with 200 mL of 0.431

Agata [3.3K]

Answer:

The answer to this question has been described in details on the screenshots attached to this question.

Thanks. Hope it helps

7 0

2 years ago

Identify the spectator ions in this reaction:<br> 2H+ + 2NO3- + Ca2+ + 2OH- → Ca2+ + 2NO3 + 2H20

inysia [295]

Answer:

Explanation:

2NO3, Ca2+

Spectator ions exists in both sides of the equation. These ions are present in both sides thus cancel out (they do not contribute to reaction...i.e. they only watch or spectate ;) )

4 0

2 years ago

Read 2 more answers

What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?

serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0

2 years ago