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3 years ago

What is the highest percentage of water vapor that could be in the air hurry help

Chemistry

1 answer:

Inessa05 [86]3 years ago

4 0

Answer:

Explanation:

At 30 °C (86 °F), for example, a volume of air can contain up to 4 percent water vapor.

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A chemist has 2. 0 mol of methanol (CH3OH). The molar mass of methanol is 32. 0 g/mol. What is the mass, in grams, of the sample

natima [27]

The mass, in grams, of the sample of methanol (CH₃OH) is 64 grams.

<h3>How we calculate mass from moles?</h3>

Mass of any substance can be calculated by using moles as:

n = W/M, where

W = required mass

M = molar mass

In the question that:

Moles of methanol = 2mole

Molar mass of methanol = 32g/mole

On putting these values in the above equation, we get

W = n × M

W = 2mole × 32g/mole = 64g

Hence, 64 grams is the mass of the sample.

To know more about moles, visit the below link:

brainly.com/question/15374113

6 0

2 years ago

A ring costs $36 more than a bracelet. the cost of the bracelet is 4/7 the cost of the ring. find the total of the two items.

avanturin [10]

So let's use some equations to represent the data [let R= cost of ring & B= cost of bracelet]

R= B + $ 36 .... (1)

B= $\frac{4}{7}$ × R ... (2)

By using simultaneous equations to solve for B and R.
Substitute eq. (1) into eq. (2)

B = $\frac{4}{7}$ × (B + $36)

B = $\frac{4}{7}$ B + $\frac{144}{7}$

$( \frac{7}{7} - \frac{4}{7} ) B = \frac{144}{7}$

$\frac{3}{7} B = \frac{144}{7}$

⇒ B = $48

By substituting value of B into ea (1)

If R = B + $36

R = ($48) + $36

= $84

∴ <span> the total of the two items = R + B
= $84 + $48
</span> = $132

7 0

3 years ago

A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2

Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

$H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O$ (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL$

Putting values in above equation, we get:

$2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL$

$M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M$

0.823 M was the molarity of the KOH solution.

7 0

2 years ago

What mass of Hg will occupy a volume of 75.0 mL?

k0ka [10]

75.0 mL in liters:

75.0 / 1000 => 0.075 L

1 mole -------------------- 22.4 L ( at STP)
( moles Hg) ------------- 0.075 L

moles Hg = 0.075 x 1 / 22.4

moles = 0.075 / 22.4

= 0.00334 moles of Hg

Hg => 200.59 u

1 mole Hg ----------------- 200.59 g
<span>0.00334 moles Hg ----- ( mass Hg )
</span>
mass Hg = 200.59 x 0.00334 / 1

mass Hg = 0.6699 / 1

= 0.6699 g of Hg

7 0

2 years ago

Explain a plastic gayer

yan [13]

Answer: seach it up

Explanation:

3 0

2 years ago