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Anna35 [415]
3 years ago
12

1. The equation for the reaction between zinc and aqueous copper(II) sulfate is shown.

Chemistry
1 answer:
Fed [463]3 years ago
3 0

Answer:

The oxidation state of the reducing agent has changed from 0 to +2.

Explanation:

reducing agent is anything that loses electron or gains oxygen

in this case, zinc

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Organs _____.
Anarel [89]

Answer:

A

Explanation:

organs are made of tissue so it can't be D.

plant are made up of organs like roots, leaves, etc. so it can't be C.

skin (is an organ) is made up to 7 layers of ectodermal tissue so isn't B.

6 0
3 years ago
An organic compound (A) reacts with HCN to give (B). On hydrolysis of (B) in acidic medium gives (C). Compound (A) also produces
Vinil7 [7]

Answer:

Hydrolic Acid

Explanation:

7 0
3 years ago
True or False: The products of a chemical
attashe74 [19]

Answer:

true

Explanation:

6 0
2 years ago
Read 2 more answers
Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and
swat32

Answer:

\%\ mass\ of\ CaCO_3=93.37\ \%

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = 293.15 K  

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.mmHg/K.mol  

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol  × 293.15 K  

⇒n of CO_2 produced =  0.0493 moles

According to the reaction:-

CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of CaCO_3 = 100.0869 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0493\ mol= \frac{Mass}{100.0869\ g/mol}

Mass_{CaCO_3}=4.93\ g

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100

\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100

\%\ mass\ of\ CaCO_3=93.37\ \%

3 0
3 years ago
Which best describes a difference between prokaryotic and eukaryotic cells
tekilochka [14]

Answer:Please find the attachment

Explanation:

7 0
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