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IRISSAK [1]
3 years ago
6

23. The compound which contains both ionic and covalent compound is

Chemistry
2 answers:
Effectus [21]3 years ago
7 0

Calcium carbonate is another example of a compound with both ionic and covalent bonds. Here calcium acts as the cation, with the carbonate species as the anion. These species share an ionic bond, while the carbon and oxygen atoms in carbonate are covalently bonded.

hoa [83]3 years ago
5 0

Answer:

D and C

Explanation:

KCl

because potassium ion ( K† ) is ionic and Chloride ion ( Cl- ) is covalent.

also KCN can be an alternative answer because the potassium ion ( K† ) is ionic and Cyanide ion ( CN- ) is covalent.

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I need help with my quiz.
katrin2010 [14]

Answer:

C

Explanation:

4 0
3 years ago
Read 2 more answers
What is the temperature -71 oC expressed in Kelvin?<br> Group of answer choices
satela [25.4K]

Answer: 202.15

Explanation:

-71°C + 273.15

= 202.15K

5 0
3 years ago
Save the ________ used to extract metals.
MrRa [10]

Answer:

Save the *minerals* used to extract metals.

6 0
2 years ago
Given the balanced equation 2KC103+ 2KC1+302
8090 [49]

is it decomp single replacement double replacement

4 0
3 years ago
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
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