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3 years ago

A gas is similar to _____because it takes up space

Chemistry

2 answers:

nydimaria [60]3 years ago

6 0

Answer:

Liquid

Explanation:

Liquid takes up space.

guajiro [1.7K]3 years ago

3 0

Answer

Liquid

Explanation:

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1.00176 x 103<br> Scientific notation

Ber [7]

Answer:

1.0318128 × 10^2

Explanation:

7 0

3 years ago

A roller coaster starts down a hill at 10 m/s. Three seconds later, its speed is 32 m/s. What is the roller coaster’s accelerati

erica [24]

Final velocity(v) = 32 m/s
Initial velocity(u) = 10 m/s

Using kinematic equation v = u + at,

32 = 10 + a(3)

32-10
--------- = a
3

a = 22
----
3

a = 7.3 m/s^2.

Hence acceleration of the roller coaster is 7.3 m/s^2.

Hope this helps!!

8 0

3 years ago

CaCo3+2HCL Cacl2+h2o+co2

Rasek [7]

Answer:

Here

Explanation:

This is an acid-base reaction (neutralization): CaCO3 is a base, HCl is an acid.

6 0

3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

3 years ago

Show all calculations. 1. 2 C 4 H 10 + 13 O 2 -> 8 CO 2 + 10 H 2 O a) what mass of O 2 will react with 400 g C 4 H 10? b) how

bazaltina [42]

$\\ \tt\hookrightarrow 2C_4H_10+13O_2\longrightarrow 8CO_2+10H_2O$

13mol of O_2 reacts with 2mols of C_4H_10.
7.5mol of O_2 reacts with 1mol of C_4H_10

No of moles:-

$\\ \tt\hookrightarrow \dfrac{Given\:mass}{Molar\:mass}$

$\\ \tt\hookrightarrow \dfrac{400}{58}$

$\\ \tt\hookrightarrow 6.89mol$

Now

Moles of O_2

$\\ \tt\hookrightarrow 7.5(6.89)=51.6mol$

Mass of O_2

$\\ \tt\hookrightarrow 51.6(32)=1651.2g$

4 0

2 years ago