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3 years ago

HELP ASAP!

1 answer:

8 0

1. Meteorologist predict the weather by using tools. They use these tools to measure atmospheric conditions that occurred in the past and present, and they apply this information to create educated guesses about the future weather. The best we can do is observe past and present atmospheric patterns and data, and apply this information to what we think will happen in the future. Meteorologists use the scientific method on a daily – and even hourly – basis!

2. They use thermometers, barometers, sling psychrometers and rain gauges. They also use anemometers, hygrometers, weather maps, weather balloons and weather satellites.

You might be interested in

How many grams of aspartame (C14H18N2O5), also known as Nutrasweet, contain 1.98×1021molecules of aspartame?

Umnica [9.8K]

Answer:

0.97 grams

Explanation:

number of molecules=no of moles × Avogadros number

number of moles=nu.of molecules ÷ 6.02×10power23 = 1.98×10power21 ÷ 6.02×10p23=0.0033 mol

mass=nu.of moles× molar mass

molar mass of aspartame=

(14×12)+18+(2× 14.0067n)+(5×16)=294.0134 u

mass = 0.0033 ×294.0134 =0.97 gram

3 0

3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago

An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam

MAXImum [283]

Explanation:

The given data is as follows.

Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence, moles of NaOH in 97 ml = $\frac{0.1090 \times 97}{1000}$

= $10.573 \times 10^{-3} mol$

Moles of HCl in 21.00 ml = $\frac{0.2060 \times 21}{1000}$

= $4.326 \times 10^{-3}$ mol

Therefore, total moles of NaOH that reacted are as follows.

$10.573 \times 10^{-3} mol$ - $4.326 \times 10^{-3} mol$

= $6.247 \times 10^{-3}$ mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

$\frac{6.247 \times 10^{-3}}{3}$

= $2.082 \times 10^{-3}$ mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

$2.082 \times 10^{-3} mol \times 40 g/mol$

= $83.293 \times 10^{-3}$ g

= 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore, 40 g NaOH = 122 g benzoic acid

So, 0.0832 g NaOH = $\frac{122 g}{40 g} \times 0.0832 g$

= 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

$\frac{0.253 g}{0.3471 g} \times 100$

= 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0

4 years ago

What is the Celsius temperature of 1 mole of a gas that has an average kinetic energy of 4,290 joules?

timofeeve [1]

Well the the answer is 70.8c but if you round it up it is 71c which I choice and got it correct so the answer is 71c

5 0

3 years ago

Find [cu2+] in a solution saturated with cu4(oh)6(so4) if [oh − ] is fixed at 2.3 ✕ 10−6 m. note that cu4(oh)6(so4) gives 1 mol

ki77a [65]

<span>

</span> $Cu_4(OH)_6(SO_4)$
<span>
You have OH- conc = </span>2.3 ✕ 10−6 m
From the formula, you can observe the ratio of Cu2+ to OH- is 4 : 6 = 2:3

So, for 2.3 ✕ 10−6 m OH-
[Cu2+] = $\frac{2}{3} \times 2.3 \times 10^{-6}$

$= 1.53 \times 10^{-6} 
$

6 0

4 years ago