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ICE Princess25 [194]
2 years ago
15

Which of the following is an example of a negative tropism?

Chemistry
1 answer:
salantis [7]2 years ago
7 0

C—stems and leaves growing upward?


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A cyclist rode at an average speed of 10 mph for 15 miles. How long was the ride?
xxMikexx [17]
If the cyclist rode at an average speed of 10mph for 15 miles...

we can solve by dividing the distance by the speed to get time using the equation...

Δspeed = Δdistance / Δtime

Δtime = Δdistance / Δspeed

Δtime = 15 miles / 10 mph = 1.5 hours
7 0
3 years ago
Read 2 more answers
Calculate the amount in grams of Na2CO3 needed in a reaction with HCl to produce 120g NaCl?
klemol [59]
The balanced chemical reaction is written as :

Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:

120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
8 0
3 years ago
Based on the periodic table why are be, bg, ca and sr in the same column/group/family?
Lesechka [4]

Answer:

The elements are in the same column/group IIA.

See the explanation below, please.

Explanation:

The elements Calcium, Strontium, Beryllium, Magnesium, Barium and Radio, belong to the group of alkaline earth metals located in group IIA of the periodic table, they require 2 electrons to complete their octet (they have 2 valence electrons). reagents than alkali metals.

7 0
3 years ago
Hi, can someone help me balance this chemistry equation:<br> H2SO4 + RbOH -&gt; Rb2SO4 + H2O
zloy xaker [14]

H2SO4 + 2RbOH -> Rb2SO4 + 2H2O

If you want an explanation, keep reading.

In the first portion, there are two hydrogen ions and four sulfate ions.

The second portion has one rubidium ions and one hydroxide ion.

On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.

Back to the right side, there is there is water (H2O).

On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).

So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).

I hope this was easy to understand.

6 0
3 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)  + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s)  + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)  + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0
2 years ago
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