When atoms are bonded together are they stable or unstable?

Explain why CaCl2 is likely to have properties similar to those of CaBr2

qaws [65]

Answer:

Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)

Explanation:

Elements are placed into different groups in the periodic table. Elements in the same group are those that have the same number of valence electrons in their outermost shell and as a result will behave similar chemically i.e. will react with other elements in the same manner.

Chlorine and Bromine are two elements belonging to group 7 of the periodic table. They are called HALOGENS and they have seven valence electrons in their outermost shell. Hence, when they form a compound with Calcium, a group two element, these compounds (CaCl2 and CaBr2) will possess similar properties because they have elements that are from the same group (halogen group).

4 0

3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

Convert 2 butanol to 2 nitrobutane

Wewaii [24]

The conversion of 2 - butanol to 2 - nitrobutane involves conversion of 2 butanol to 2 - bromobutane and subsequently to 2 - nitrobutane.

In order to convert 2 - butanol to 2 - nitrobutane, an SN2 reaction is first carried out in which 2 - butanol is converted to 2 - bromobutane. This reaction proceeds with inversion of configuration at the chiral carbon as expected.

This product is subsequently reacted with silver nitrite in ethanol to yield 2 - nitrobutane along side silver bromide. Some alkyl nitrite is also produced as a by product. The components of the mixture are separated by fractional distillation. The scheme of the reaction is shown in the image attached to this answer. This reaction is only applied to primary and secondary alkyl halides.

Learn more: brainly.com/question/10079361

3 0

3 years ago

A large pot (10 kg) of beef stew is left out on the kitchen counter to cool. The rate of cooling, in kJ/min, equals 1.955 (T-25)

olga_2 [115]

Answer:

16 minutes

Explanation:

First, we need to calculate the amount of heat needed to cool the beef stew:

Q = mcΔT

Where m is the mass, c is the heat capacity and ΔT is the variation of the temperature.

Q = 10x4x(40 - 90)

Q = -2000 kJ

So, the beef stew needs to lost 2000 kJ to cool.

With the initial temperature at 90ºC, the rate of cooling(r) will be:

r = 1.955x(90 - 25)

r = 127.075 kJ/min

So, to lose 2000 kJ, will be necessary:

t = Q/r

t = 2000/127.075

t = 16 minutes

5 0

3 years ago

Why don't I need prefixes to designate the number of each type of element when naming ionic compounds?

scoray [572]

We use prefixes in ionic compounds Don't use numeric prefixes like mono, di, tri, etc. when naming ionic compounds - they are used only to denote covalent molecular compounds.

5 0

2 years ago