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3 years ago

15

In science class, Jake mixed water with differing amounts of an unknown liquid. After mixing the liquids, he heated 20 millilite

rs (ml) of each mixture and observed how quickly it boiled. The table shows his results.

1 answer:

Snowcat [4.5K]3 years ago

7 0

Answer:

decrease i had the same question in 8th grade u must be in ninth

Explanation:

ohh im not sure but i think its A

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A reaction where one becomes two

2H202 --> 2H2O + O2

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2 years ago

An organic solvent has a density of 1.31 g/mL. What volume is occupied by 57.5 g of the liquid?

Burka [1]

Answer:

<h3>The answer is 43.89 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question

mass = 57.5 g

density = 1.31 g/mL

We have

$volume = \frac{57.5}{1.31} \\ = 43.89312977...$

We have the final answer as

<h3>43.89 mL</h3>

Hope this helps you

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2 years ago

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

<span>We have the following statement data:
</span>
Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

<span>As the percentage is the mole fraction multiplied by 100:

</span> $P = X_{ O_{2} }*100$

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture?
</span>
$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$
<span>
Answer:
</span><span>b. 320.0 mm hg </span>

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3 years ago

A titration of 0.1 M NaOH into 1.2 L of HCl was stopped once the pH reached 7 (at 25C). If 0.4 L of NaOH needed to be added to a

MArishka [77]

Answer:

0.033 M

Explanation:

Let's consider the neutralization reaction between NaOH and HCl.

NaOH + HCl → NaCl + H₂O

0.4 L of 0.1 M NaOH were used. The reacting moles of NaOH are:

0.4 L × 0.1 mol/L = 0.04 mol

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 0.04 moles.

0.04 moles of HCl are in 1.2 L. The molarity of HCl is:

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