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AlladinOne [14]

3 years ago

15

A neutral solution has a pH of: a. 1 b. 5 c. 7 d. 14

1 answer:

zavuch27 [327]3 years ago

3 0

The answer is 7 my guy

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The mass of a single gold atom is 3.27X10^-22 grams. How many gold Adams with there be in 57.8 mg of gold.

vichka [17]

Answer:

18 * 10^19 atoms

Explanation:

We must first convert 57.8 mg to grams.

If 1000 mg = 1g

57.8 mg = 57.8/1000 = 57.8 * 10^-3 g

Now;

If 1 gold atom has a mass of 3.27X10^-22 grams

x gold atoms have a mass of 57.8 * 10^-3 g

x = 57.8 * 10^-3 g/3.27X10^-22 g

x = 18 * 10^19 atoms

7 0

3 years ago

A gas at 325 k has a volume of 4.0 L. If the gas’s volume changes to 2.0 L at constant pressure, what will be the new temperatur

inessss [21]

4/325 = 2/unknown temperature
unknown temperature= 2/(4/325)=162.5k

8 0

4 years ago

Read 2 more answers

The specific heat of copper is 0.40 joules/ g °c. How much heat is needed in joules to change the temperature of a 55 gram subst

Snezhnost [94]

Answer : The amount of heat needed is, 1188 J

Explanation :

Formula used :

$q=m\times c\times (T_2-T_1)$

where,

q = heat needed = ?

m = mass of copper = 55 g

c = specific heat capacity of copper = $0.40J/g^oC$

$T_1$ = initial temperature = $20.0^oC$

$T_2$ = final temperature = $74.0^oC$

Now put all the given values in the above formula, we get:

$q=55g\times 0.40J/g^oC\times (74.0-20.0)^oC$

$q=1188J$

Thus, the amount of heat needed is, 1188 J

3 0

4 years ago

How many control(s) are in an experiment

weqwewe [10]

You can have as many controls as necessary, But they must remain equal at all times in order to get the most accurate results

6 0

3 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

4 years ago