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blagie [28]
2 years ago
10

A student is trying to classify an unidentified, solid gray material as a metal or a nonmetal. Which question will best help the

student classify the material?.
Chemistry
2 answers:
guajiro [1.7K]2 years ago
8 0

The question that will best help the student to classify the material is; "is the material malleable or ductile?"

Generally, materials can be classified as metals or non metals. There are properties that are particular to metals and there are properties that are particular to nonmetals and these properties can be used to identify each one of the materials.

The question that will best help the student to classify the material is; "is the material malleable or ductile?" These metallic properties.

Learn more:

brainly.com/question/1659592

Missing parts;

A student is trying to classify an unidentified, solid gray material as a metal or a nonmetal. Which question will best help the student classify the material? A. Is the material malleable or ductile? B. Does the material feel hard to the touch? C. Will the material float in water? D. Does the material feel rough or smooth?

Kruka [31]2 years ago
5 0
Whether it conducts heat or electricity,if it is attracted by a magnet , high density high melting point
You might be interested in
The combined gas law states that
seropon [69]

Answer:

Option B. PV = nRT.

Explanation:

From the above,

C = PV / T. (1)

But, C is directly proportional to n, where R is the gas constant i.e

C & n

C = nR

Substituting the value of C into equation (1), we have:

C = PV / T

nR = PV / T

Cross multiply to express in linear form

nRT = PV

Therefore,

PV = nRT

8 0
2 years ago
Please help me with these
padilas [110]
<span>The physical layers of the inner core and outer core form a single compositional layer.

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8 0
2 years ago
A Review Constants Periodic Table
soldi70 [24.7K]

Answer:

C. 1, 4–dimethyl–1–cyclohexene

D. 4, 6–dichloro–1–heptene

Explanation:

C. Determination of the name of the compound.

To name the compound given above we must observe the following:

1. Determine the functional group of the compound. In this case, the functional group is the double bond (–C=C–).

2. Determine the parent name by counting the number of carbon that makes up the ring. In this case, 6 carbon makes up the ring. Hence the parent name is cyclohexene.

3. Identify the substituent group attached to the compound. In this case, two methyl group i.e –CH₃ are attached.

4. Locate the position of the substituent group attached by giving the functional group the lowest possible count. In this case, the functional group is at carbon 1, the first methyl group is at carbon 1 and the 2nd methyl group is at carbon 4.

5. Combine the above to obtain the name. Thus, the name of the compound is:

1, 4–dimethyl–1–cyclohexene.

D. Determination of the name of the compound.

To name the compound given above we must observe the following:

1. Determine the functional group of the compound. In this case, the functional group is the double bond (–C=C–).

2. Determine the parent name by counting the number of carbon that makes up the chain. In this case, 7 carbons makes up the chain. Hence, the parent name of the compound is heptene.

3. Identify the substituent group attached to the compound. In this case, two chlorine (Chloro) i.e –Cl are attached.

4. Locate the position of the substituent group attached by giving the functional group the lowest possible count. In this case, the functional group is at carbon 1, the first chlorine is at carbon 4 and the 2nd is at carbon 6.

5. Combine the above to obtain the name. Thus, the name of the compound is:

4, 6–dichloro–1–heptene

8 0
2 years ago
If i initially have 4.0 l of a gas at a pressure of 1.1 atm, what will the volume be if i increase the pressure of 3.4 atm?
ddd [48]
  The  volume  of  a  gas  that   its  pressure  increase  to  3.4  atm   is    calculated  as   follows

  By  use  of  boyles   law   that  is  P1V1=P2V2
V1=4.0  L
P1=1.1  atm
P2=3.4  atm
V2= P1V1/P2  

(1.1  atm  x  4.0 L)/3.4  atm=  1.29  L
4 0
2 years ago
Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t
myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol

Moles of methane gas = n_1=\frac{8.00 g}{16 g/mol}=0.5 mol

Moles of ethane gas  =n_2=\frac{18.0 g}{30 g/mol}=0.6 mol

Moles of propane gas = n_3

n=n_1+n_2+n_3

2.22=0.5 mol +0.6 mol+ n_3

n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol

Mole fraction of methane =\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}

\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252

Similarly, mole fraction of ethane and propane :

\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703

\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

Partial pressure of methane gas:

p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm

Partial pressure of ethane gas:

p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm

Partial pressure of propane gas:

p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = n_1=\frac{37.8 g g}{28g/mol}=1.35 mol

Moles of oxygen gas  =n_2=\frac{62.2 g}{32 g/mol}=1.94 mol

Mole fraction of nitrogen=\chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103

Similarly, mole fraction of oxygen

\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg

Partial pressure of oxygen gas:

p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg

3 0
3 years ago
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