All categories

3 years ago

PLEASE HELP! I will give brainliest

Chemistry

1 answer:

horsena [70]3 years ago

6 0

Answer:

3.07g of NaCl are present in the solution

Explanation:

A 1.05 M NaCl solution means you have 1.05 moles of NaCl per liter of solution. In 50.0mL = 0.0500L you will have:

0.0500L * (1.05mol / L) = 0.0525 moles of NaCl.

To convert these moles to grams you must use molar mass of the compound (Molar mass NaCl: 58.44g/mol). In 0.0525 moles of NaCl you will have:

0.0525mol * (58.44g / mol) =

<h3>3.07g of NaCl are present in the solution</h3>

You might be interested in

a sample of water with a mass of 648.00 kg at 298 K is heated with 87 kh of energy. the specific heat of water is 1 J-1 kg K-1.

Stels [109]

q = mCΔT

The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.

ΔT = q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K

Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K

3 0

3 years ago

Read 2 more answers

How many moles are in 1.20 x 10^21 atoms of Phosphorus?

Ludmilka [50]

Answer:

You would get 19.969 moles

Explanation:

8 0

3 years ago

A hot object will have ___ kinetic energy

Masja [62]

Answer:

More/ Alot? I think is what you are looking for?

Explanation:

It will definitely have some but I'm not sure on what word you are looking for.

6 0

3 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

$n=\frac{m}{M_m}$

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

$n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol$

Now we can find the pressure of the gas by using the ideal gas law:

$pV=nRT$

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

$p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm$

8 0

4 years ago

A 0.98 gram sample of a volatile liquid was heated to 348 k. the gas occupied 265 ml of space at a pressure of 0.95 atm. what is

arlik [135]

Answer:

The molecular weight is $Z = 111.2 \ g/mol$

Explanation:

From the question we are told that

The mass of the sample is $m = 0.98 \ g$

The temperature is $T = 348 K$

The volume which the gas occupied is $V = 265 \ ml = 265 *10^{-3} L$

The pressure is $P = 0.95 \ atm$

Generally from the ideal gas equation we have that

$PV = n RT$

Here n is the number of moles of the gas while the R is the gas constant with value $R = 0.0821 \ atm \cdot L \cdot mol^{-1} \cdot K^{-1}$

$n = \frac{PV}{ RT}$

=> $n = \frac{ 0.95 * 265 *10^{-3} }{ 0.0821 * 348}$

=> $n = 0.00881 \ mol$

Generally the molecular weight is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{0.98 }{0.00881}$

=> $Z = 111.2 \ g/mol$

8 0

3 years ago