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Alinara [238K]
3 years ago
8

A base is a substance that accepts protons. true or false?

Chemistry
2 answers:
oee [108]3 years ago
5 0

Answer:

true

Explanation:i took a test got it right

umka21 [38]3 years ago
3 0
True................
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What is the ph of a peach with a [ –oh] = 3.2 × 10 –11 m?
3241004551 [841]

Answer:

Amerigo Vespucci, was the first European to reach the Caribbean Islands.

Please select the best answer from the choices provided

T

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Explanation:

7 0
2 years ago
What is the relationship between frequency and wavelength?
IrinaVladis [17]

Answer:

frequency it the measure of the wave length. The measure of the peaks and troughs is how you measure the frequency. the distance between these is the wave length.

6 0
3 years ago
Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g
Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water  = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

\frac{Mass}{Volume}

Mass = \frac{percent  \;of  \;ethanol  \;X  \;density  \;of  \;ethanol  \;+  \\ \;percent  \;of  \;water X  \;density  \;of  \;water}{100}

0.926 = \frac{x X  0.7890 g/mL  \;+  (100-x) X  0.9982 g/mL}{100}

x= 34.51 %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

Learn more about the density here:

brainly.com/question/952755

#SPJ1

8 0
2 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?
slava [35]

Answer:

A. 0.90 L.

Explanation:

  • NaOH solution will react with H₂SO₄ according to the balanced reaction:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>

<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>

  • For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>

4 0
3 years ago
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