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MakcuM [25]
3 years ago
5

Need help fast !!!!!!!

Chemistry
2 answers:
Sveta_85 [38]3 years ago
6 0

Answer:

Brittle

Explanation:

because thas for non metals

Oduvanchick [21]3 years ago
6 0
The answer is the last choice which is brittleness
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Balance :FeCl3 + __Ca(OH)2 → ___ Fe(OH)3 + __ CaCl2
ArbitrLikvidat [17]
Your answer is 2:3:2:3
7 0
3 years ago
If a 275 mL gas container had pressure of 732.6 mm Hg at -28°C and the gas was condensed into a liquid with a mass of 1.95 g, wh
ipn [44]

Answer:

THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE

Explanation:

Using PV = nRT

n = Mass / molar mass

P = 732.6 mmHg = 1 atm = 760 mmHg

So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm

P = 0.964 atm

V = 275 mL = 275 *10 ^-3 L

R = 0.082 Latm/ mol K

T = -28 C = 273 - 28 K = 245 K

mass =  1.95 g

molar mass = unknown

Having known the other variables in the formula, the molar mass of the gas can be obtained.

PV = m R T/ molar mass

Molar mass = m RT / PV

Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3

Molar mass = 39.1755 / 265.1 *10^-3

Molar mass = 39.1755 / 0.2651

Molar mass = 147.78 g/mol

The molar mass of the gas is 147.78 g/mol

5 0
4 years ago
Is it true dalton inferred that all atoms of an element are identical?
Olegator [25]
Yes and no he said that all atoms of a given element are identical in mass and properties.
7 0
3 years ago
The atomic mass of sulfur is 32.1 amu, and the atomic mass of oxygen is 16.0 amu. To the nearest tenth of a percent, what is the
Zielflug [23.3K]

Answer:

\%\ Composition\ of\ sulfur=40.1\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given , Mass of sulfur= 32.1 amu

Mass of oxygen = 16.0 amu

Mass of sulfur trioxide SO_3 = 32.1 amu + 3*16.0 amu = 80.1 amu

\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100

\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%

6 0
3 years ago
9. Let's say, you need to make 2.00 L of 0.05 M copper (II) nitrate solution. How many grams of
Vlad [161]

Answer:

18.76 g of copper II nitrate

Explanation:

Now recall that we must use the formula;

n= CV

Where;

n= number of moles of copper II nitrate solid

C= concentration of copper II nitrate solution

V= volume of copper II nitrate solution

Note that;

n= m/M

Where;

m= mass of solid copper II nitrate

M= molar mass of copper II nitrate

Thus;

m/M= CV

C= 0.05 M

V= 2.00 L

M= 187.56 g/mol

m= the unknown

Substituting values;

m/ 187.56 g/mol = 0.05 M × 2.00 L

m= 0.05 M × 2.00 L × 187.56 g/mol

m= 18.76 g of copper II nitrate

Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.

6 0
3 years ago
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