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3 years ago

HHHEEEEEEELLLLPPPP PPPPPLLLLEEEASSSEEEE I GIVE ~BRAINLIEST~!!!!!! PLZ HELP ME

Chemistry

2 answers:

AnnyKZ [126]3 years ago

5 0

The action or power of focusing one's attention or mental effort.

a substance produced by a living organism that acts as a catalyst to bring about a specific biochemical reaction.

a substance that slows down or prevents a particular chemical reaction or other process, or that reduces the activity of a particular reactant, catalyst, or enzyme.

Anni [7]3 years ago

4 0

Concentration-The attention or power of focusing one's attention of mental effort.
Enzyme-Substance produced by living organism that acts as a catalyst to bring about a specific biochemical reaction
Inhibitor-Things that inhibits someone or something

You might be interested in

Acidic Basic Neutral

lana66690 [7]

The answer is acidic

8 0

3 years ago

Identify the spectator ions in the reaction that occurs between aqueous solutions of potassium hydroxide and hydrochloric acid

m_a_m_a [10]

Answer:

The answer is option A.

K+ Cl-

Hope this helps you

8 0

3 years ago

Using the unbalanced reaction below:

Vladimir [108]

1.16 g of UF₄ can be formed in this reaction.

Explanation:

We have to write the balanced equation of the reaction between Uranium oxide and hydrogen fluoride to form Uranium tetra fluoride and water as,

UO₂(s) +4 HF(aq) → UF₄ + 4 H₂O(l)

We can find the moles of UF₄ and then multiplying molar mass of UF₄ with the moles of UF₄, we will get the mass of UF₄ in grams.

Moles of UF₄ = $$ \frac{ 1 mol of UO_{2}\times 1 mol of UO_{2\times 1 mol of UF_{4} } }{270.03 g \times 1 mol of UO_{2} }$

= 0.00370 mol UF₄

Grams of UF₄ =0.00370 mol × 314.02 g/mol = 1.16 g of UF₄

So mass of UF₄ formed in the reaction is 1.16 g.

5 0

4 years ago

2. When Thomson held a positively charged plate near the cathode ray, the beam bent toward the plate. What conclusion can be dra

loris [4]

Answer:

option B is correct : The beam was positively charged.

Explanation:

In 19th century J.J Thomson conducted an experiment on cathode ray tube. These cathode ray tubes were sealed glass tubes inside which the pressure of the gas reduced by evacuating air. Then he applied a high voltage across two electrodes at end of the tube, that voltage caused a beam of particles to flow from the cathode to the anode.

Beam of the particles were originated at the cathode and detected on anode by a phosphorous beyond the anode. As at that time phosphorous sparks and emit light.

To test properties of particles he places two opposite charged electric plates around the cathode rays that is one negative and one positive plate.

Thomson observed that these rays deflected to positive electric plate so it indicated that these rays are composed of negatively charged particles as it was attracted by the positive charged electric plate. These particles later named as electrons.

So,

option B is correct : The beam was positively charged.

8 0

3 years ago

Stoichiometry. Are my answers right?

OverLord2011 [107]

Answer:

Congratulations. Mostly correct.

Step-by-step explanation:

Part I. Gold

Moles of Au = 35.12 2g × 1/196.97 = 0.1783 mol Au

Atoms of Au = 0.1783 × 6.022 × 10²³/1 = 1.074 × 10²³ atoms Au

Part II. Sucrose

Molar mass = 342.30 g/mol

Moles of C₁₂H₂₂O₁₁ = 1.202 g × 1/342.30

Moles of C₁₂H₂₂O₁₁ = 0.003 512 mol C₁₂H₂₂O₁₁

Moles of C = 0.003 512 × 12/1 = 0.042 14 mol C

Moles of H = 0.003 512 × 22/1 = 0.077 25 mol H

Moles of O = 0.003 512 × 11/1 = 0.038 63 mol O

Atoms of C = 0.042 14 × 6.022 × 10²³ = 2.538 × 10²² atoms C

Atoms of H = 0.077 25 × 6.022 × 10²³ = 4.652 × 10²² atoms H

Atoms of O = 0.038 63 × 6.022 × 10²³ = 2.326 × 10²² atoms O

Minor Quibbles:

You put the moles of sucrose in the molar mass slot.

You got the wrong exponents for the atoms of C, H, O.

You followed the rules for significant figures for gold, but not for sucrose.

6 0

3 years ago