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2 years ago

Name an acids and a bases

Chemistry

1 answer:

Bumek [7]2 years ago

6 0

Answer:

HCl for acid, NaOH for base.

Explanation:

Acids start with H.

Bases end with OH.

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What is the mass of an object with a density of 3.4 g/mL and a volume of 500.0 mL

umka2103 [35]

Formula:
$mass=density *volume$

Given:
Density=3.4
Volume=500.0

Plug them into the formula:
$mass=3.4 *500.0=1700$

Final answer: 1700g

5 0

2 years ago

Select the correct answer

vekshin1

Answer:

Temperature

Explanation:

Kinetic energy of gass molecules is directly propotional to the temperature.

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2 years ago

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Name the process by which the ammonia and hydrogen Chlorade gass nove in the tube-

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Explanation: Process is diffusion. The gas molecules follow a zig-zag path through the tube as they collide with the air molecules in the tube. The purpose of the glass tube is to eliminate air currents and to see if the gas molecules will move on their own.

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2 years ago

This week in lab you will use the spectrovis plus spectrophotometer to measure the light that is absorbed by dopachrome, a dark

NARA [144]

A solution that appears dark orange in color is not absorbing all orange wavelengths of light.

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3 years ago

What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium

alex41 [277]

Answer:

Approximately $3.06 \times 10^{2}\; \rm g$ (approximately $306\; \rm g$ .)

Explanation:

Calculate the quantity $n$ of lithium phosphate in $V = 2.5\; \rm L$ of this $c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ lithium phosphate solution.

$\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}$ .

Empirical formula of lithium phosphate: $\rm Li_3PO_4$ .

Look up the relative atomic mass of $\rm Li$ , $\rm P$ ,and $\rm O$ on a modern periodic table:

$\rm Li$ : $6.94$ .
$\rm P$ : $30.974$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm Li_3PO_4$ :

$M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}$ .

Calculate the mass of that $n = 2.65\; \rm mol$ of $\rm Li_3PO_4$ formula units:

$\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}$ .

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2 years ago