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2 years ago

Which two tools are used to measure humidity?

Chemistry

2 answers:

Rainbow [258]2 years ago

8 0

Answer:

Explanation:

This is because barometer is used to measure pressure

dsp732 years ago

3 0

Answer:

B they are bo thg basically the same

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What is the volume of 40g of sugar given it’s listed density

lora16 [44]

Answer:

25.157 cm³

Explanation:

Data Given:

Mass of Sugar (m) = 40g

Density of sugar given in literature = 1.59 g/cm³

Volume of Sugar = ?

The formula will be used is

d = m/v ........................................... (1)

where

D is density

m is the mass

v is the volume

Rearrange the Equation (1)

d x v = m

v = m/ d ................................................ (2)

put the given values in Equation (2)

v = 40g / 1.59 g/cm³

v = 25.157 cm³

volume of 40 g of sugar = 25.157 cm³

8 0

2 years ago

A scientist experiments ima unknown solution. She notices that when a base, sodium hydroxide, is added to the solution, it forms

Lelechka [254]

The unknown solution is acidic

8 0

2 years ago

Read 2 more answers

I need your help!! about Chemistry (15 point!!)

irakobra [83]

1. I think it is true?
2. Low melting points
3. True
4. Atomic number, I think it’s periods?
5. Groups?

Sorry, I might not get all of them right :(

Hope this helps you in any way!!

6 0

2 years ago

Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for M₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

7 0

2 years ago

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

klemol [59]

Let us start with the total area of the lawn. Area= width x length, ie, 21 x 20 = 420 sq. ft. Snow flakes per square foot per minute = 1350 So Snow flakes for 420 sq.feet per minute = 420 x 1350 = 567000. Snow flakes for 1 hour = 567000 x 60 = 34020000 (60 minutes) Weight of 34020000 snow flakes = 34020000 x 1.60 = 54432000mg. To convert it into kilograms, divide this number by 1000000 (1 kilogram = 1000000 milligrams) Thus 54432000/1000000 = 54.432 kilograms or 54 kilograms and 432 grams.

6 0

2 years ago