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3 years ago

Can someone please check my answer??!!!

Chemistry

2 answers:

Ksenya-84 [330]3 years ago

5 0

Answer:

I think its correct

Vika [28.1K]3 years ago

3 0

The two units cancel out.

Like a fraction, when the same terms are in the top and bottom, they can be cancelled. Since the same units are in the numerator and denominator, they can be cancelled out.

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vladimir2022 [97]

Answer:

A. glass and diamond

Explanation:

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3 years ago

How do you determine the amplitude of a transverse wave

barxatty [35]

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3 years ago

Read 2 more answers

How many moles are in 1.42 x 10^25<br> molecules of NaCl?

Sholpan [36]

<h3>Answer:</h3>

23.6 mol NaCl

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.42 × 10²⁵ molecules NaCl

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.42 \cdot 10^{25} \ molecules \ NaCl(\frac{1 \ mol \ NaCl}{6.022 \cdot 10^{23} \ molecules \ NaCl})$
Multiply/Divide: $\displaystyle 23.5802 \ mol \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

23.5802 mol NaCl ≈ 23.6 mol NaCl

3 0

2 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 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

3 years ago