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yan [13]
3 years ago
10

Name this compound Ca (OH)2

Chemistry
2 answers:
Ira Lisetskai [31]3 years ago
7 0

I believe the answer is Calcium dihydroxide

devlian [24]3 years ago
3 0

Your answer is calcium hydroxide

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Which of the following are physical characteristics of blood? 1. it is more viscous than water 2. its temperature is about 38ºC
yawa3891 [41]

Explanation:

A property which does not bring any change in chemical composition of a substance is known as physical property.

For example, blood is more viscous than water, its pH is slightly alkaline, its temperature is about 38^{o}C (100.4^{o}F)  .

On the other hand, a property which changes chemical composition of a substance is known as a chemical property.

For example, precipitation, reactivity, toxicity etc are chemical property.

Sometimes, the change in color of a substance can also occur due to a chemical change.

Thus, we can conclude that following are the physical characteristics of blood.

  • it is more viscous than water.
  • its temperature is about 38^{o}C (100.4^{o}F).
  • the pH is slightly alkaline.
  • it is about 8% of one’s total body weight.
6 0
3 years ago
Calculate the molar mass of a gas
spin [16.1K]

Answer:

MOLAR MASS = 32 g/mol

Explanation:

Condition of standard temperature and pressure(STP) are as follow:

Temperature = 273 K

Pressure = 1 atm (or 100000 Pa)

Here atm is atmosphere and Pa is Pascal

STP conditions arte used for measuring gas density and volume using Ideal Gas Law.Here 1 mole of ideal gas occupies 22.4 L of volume.

According toi Ideal Gas Equation :

PV = nRT

where P = pressure, n= number of moles, V = volume ,R= Ideal Gas Constant and T= temperature

n=\frac{PV}{RT}

From question:

V=280 ml = 0.28 L

P = 1 atm

R=0.08205 L atm/K mol

T=273 K

Putting values in above formula :

n=\frac{1\times .280}{0.08205\times 273}

n = 0.0125 moles

Now n=\frac{given\ mass}{Molar\ mass}

Molar\ mass=\frac{given\ mass}{n}

given mass = 0.4 g (given)

Molar\ mass=\frac{0.4}{0.0125}

On solving we get:

Molar mass = 32 g/mol

4 0
3 years ago
he Lewis structure for CO molecule contains Group of answer choices one double bond, one single bond, and twelve nonbonding elec
zimovet [89]

Answer:

One triple bond and four non bonding electrons

Explanation:

In considering the lewis structure of carbon monoxide, we must remember that the molecule contains a total of ten valence electrons. Four are the valence electrons that are present on the valence shell of carbon while six are the valence electrons on oxygen. Some of these valence electrons participate in bonding in the CO molecule.

Out of the six valence electrons on oxygen, two valence electrons participate in bonding with carbon while the other four electrons remain localized on the oxygen atom as two lone pairs of electrons.

Hence there are four nonbonding electrons in the lewis structure of CO as well as one triple bond.

6 0
4 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
Given the speed of light as 3.00 x 10° m/s, calculate the
Anarel [89]

Answer:

the answer is mabye 30 I didnt understand

6 0
2 years ago
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