A sample of water at 20 degrees c contains what bonds

What is the molar concentration of the acid if 35.18 mL of hydrochloric acid was required to neutralize 0.745 g of ALUMINUM hydr

Makovka662 [10]

The mass number of aluminium hydroxide is 78 thus, the number of moles in 0.745 g is:

no. of moles= mass/ RFM

= 0.745/78

=0.00955moles

Therefore the 0.00955 moles should be in the 35.18 ml

therefore 1000ml of the solution will have:

(0.00955ml×1000ml)/35.18

=0.2715moles

The solution will be 0.27M hydrochloric acid

3 0

3 years ago

Need help asap!!! (SCIENCE)

balu736 [363]

<h3><u>Answer;</u></h3>

10 percent

<h3><u>Explanation;</u></h3>

Probability is a number that describes how likely it is that an event will occur.

In this case; If each of 10 events is equally likely to occur, the probability of each individual event occurring is 1/10.

Therefore; as a percentage;

1/10 × 100 = 10 percent

Hence; The probability of each individual event occurring is 10 percent

6 0

3 years ago

. How many ethyl groups are in the molecule 3, 3-diethylpentane?

Mamont248 [21]

Answer:

2 groups

Explanation:

The molecule 3, 3-diethylpentane has 2 ethyl groups because the prefix di- means that there are 2 homogeneous substituent groups present in the molecule.

6 0

2 years ago

. Monthly measurements of atmospheric CO2 concentration at Mauna Loa began in March 1958. The average CO2 concentration for that

Harrizon [31]

Answer:

$Increase=98.8ppm$

$Average\ increase/year=1.594\frac{ppm}{year}$

Explanation:

Hello,

In this case, since the nowadays concentration of CO2 is 414.51 ppm and the concentration in 1958 was 315.71 ppm, the total increase is computed via the difference between them:

$Increase=414.51ppm-315.71ppm\\\\Increase=98.8ppm$

Moreover, the average increase per year is computed considering that from 1958 to 2020, 62 years have passed, therefore, such average is:

$Average\ increase/year=\frac{98.8ppm}{62 years} \\\\Average\ increase/year=1.594\frac{ppm}{year}$

Regards.

4 0

3 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

4 years ago