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kondor19780726 [428]
3 years ago
6

When elements combine to form a mixture, the resulting chemical properties may be very different from those of the elements that

make it. true or false
Chemistry
2 answers:
vladimir2022 [97]3 years ago
6 0

Answer:

True

Explanation:

guapka [62]3 years ago
5 0
True i really don’t feel like giving an explanation but trust me it’s trye
You might be interested in
What is an atom? please answer ASAP
Pani-rosa [81]

Answer:

smallest particle of an element which can take part in a chemical reaction

Explanation:

there are atoms of different elements is sodium atom

3 0
3 years ago
The water in a 2500-L aquarium contains 1.25 g of copper. Calculate the concentration of copper in the water in ppm (remember 1
lorasvet [3.4K]

Answer:

0.5ppm

Explanation:

Step 1:

Data obtained from the question.

Volume of water = 2500L

Mas of Cu = 1.25 g

Step 2:

Determination of the concentration of Cu in g/L. This is illustrated below:

Volume of water = 2500L

Mas of Cu = 1.25 g

Conc. of Cu In g/L =?

Conc. g/L = Mass /volume

Conc. of Cu in g/L = 1.25/2500

Conc. of Cu in g/L = 5x10^–4 g/L

Step 3:

Conversion of the concentration of Cu in g/L to ppm. This is illustrated below

Recall:

1g/L = 1000mg/L

Therefore, 5x10^–4 g/L = 5x10^–4 x 1000 = 0.5mg/L

Now, we know that 1mg/L is equal to 1ppm.

Therefore, 0.5mg/L is equivalent to 0.5ppm

6 0
3 years ago
A piece of unknown metal weighs 348g. When the metal piece absorbs 6.64kj of heat , its temperature increases from 24.4C to 43.6
Morgarella [4.7K]

Answer:

This metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

Explanation:

A pie of unknown metal presents a mass (M) of 348 g. This metal is heated using energy (E) of 6.64 kJ and the temperature increases from T1 =24.4 to T2 =43.6°C. We can calculate the specific heat (H) of this metal as follows

H = \frac{E}{M*(T2-T1)}

We can replace previously presented data in this equation. After simplifying and converting to adequated units, we found that

H = \frac{6640 J}{0.348 Kg*(43.6-24.4) C} =\frac{6640 J}{6.686 KgC}

Finally, the specific heat of this metal is

H = 993 \frac{J}{kgC}

The aluminium could be the metal, its specific heat is similar to that found in this problem.

Finally,  we can conclude that this metal could be the aluminium with a specific heat of H = 993 \frac{J}{kgC}

7 0
3 years ago
What is the pOH of a<br> 5.6 x 10-5 M solution of cesium<br> hydroxide (CsOH)?
Rudiy27

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

7 0
3 years ago
The pressure on 30 milliliters of an ideal gas increases from
Nana76 [90]

V1 = 30 mL

P1 = 760 torr

P2 = 1520 torr

V2 = ?

applying Boyle's Law

P1*V1 = P2*V2

760 torr * 30 mL = 1520 torr * V2

V2 = 760 torr * 30 mL / 1520 torr

( C ) is correct

8 0
3 years ago
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