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irga5000 [103]
4 years ago
12

Which of the following shows 3x + 15 + 6x – 7 + y in simplest terms?

Chemistry
1 answer:
prisoha [69]4 years ago
3 0

Answer:

9x+y+8

Explanation:

We know this because you add 3x and 6x to get 9x

then we only have one y so put y by itself then subtract 15 and y which gets us 8

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V125BC [204]

Answer:

sun wind water

Explanation:

3 0
3 years ago
It’s is the same one from my recent ones
UkoKoshka [18]
<h2>9. Given : P + O2 = P2O5</h2>

Balanced Chemical Equation: <u>4</u>P + <u>5</u>O2 = <u>2</u>P2O5

<u>Reactant Product</u>

P = (1) 4 | P = (2) 4

O = (2) 10 | O = (5) 10

Reaction Information

Phosphorus + Dioxygen = <u>Phosphorus Pentoxide</u>

Reaction Type: <u>Synthesis</u>

<h2>10. Given: HCl + NaOH = NaCl + H2O </h2>

<u>Reactant Product</u>

H = 2 | H = 2

Cl = 1 | Cl = 1

Na = 1 | Na = 1

O = 1 | O = 1

<h3>Chemical Equation is already in balance.</h3>

Reaction Information<em>:</em><em> </em><em>Hydrogen Chloride </em>+ <em>Sodium Hydroxide</em> = <em>Sodium Chloride </em>+ <em>Water</em>

Reaction Type<u> </u><u>:</u><u> </u><u>Double Displacement (Acid-Base)</u>

<h3>For more info:</h3>

<u>How to balance chemical equation?</u>

brainly.com/question/15939935

5 0
4 years ago
Which three factors are most important in determining the composition of ocean water
AfilCa [17]
First factor: Density of water
This density of water is affected by the level of salt present in it. They are inversely proportional. This means that as the amount of salt in water increases (water becomes more salty), the density of water would decrease

Second factor: Temperature
Temperature affects the composition of water because as temperature increases, the ocean water evaporates and the salt precipitates. Therefore, this factor is essential in determining the rate by which the ocean water evaporates

Third factor: Salinity
This factor helps in determining the amount of salt present in the ocean, i.e, concentration of salt in the water.

Hope this helps :)


4 0
3 years ago
Read 2 more answers
Ricky and Shemane, who are stoked to be learning stoich, were working diligently on a chemistry assignment in class. Ricky reali
kondaur [170]
38373737267383838282828228)
6 0
3 years ago
Calculate the pH during the titration of 30.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 29.3 mL of the base have be
pashok25 [27]

Answer:

3.336.

Explanation:

<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>

<em />

So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>

<em></em>

<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>

∵ pH = - log[H⁺].

<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>

7 0
4 years ago
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