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3 years ago

How many grams are there in 457.25 pounds?

Chemistry

2 answers:

Usimov [2.4K]3 years ago

8 0

there are 2.2 lbs /kg. take 457.25/2.2=207.8409091
there are 1000g/kg so take your answer times 1000= 207840.9091g

Ne4ueva [31]3 years ago

7 0

There are 207405.111 grams in that many pounds.

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Generally, most water reabsorption in a nephron occurs at the:_________

Allisa [31]

Answer:

proximal convoluted tubule and is called obligatory water reabsorption.

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3 years ago

100 POINTS PLSSS HELPPP

VARVARA [1.3K]

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3 years ago

What is the molarity of 2.3 mol of Kl dissolved in 0.5 L of water

Nastasia [14]

Answer:

$4.6\,\,moL\,\,L^{-1}$

Explanation:

Molarity refers to a measure of concentration.

Molarity = moles of solute/Litres of solution

Molarity refers to number of moles of solute present in this solution.

In order to find a solution's molarity, use value for the number of moles of solute and the total volume of the solution expressed in liters

As molarity of 2.3 mol of Kl is dissolved in 0.5 L of water,

Molarity = $\frac{2.3}{0.5} =4.6\,\,moL\,\,L^{-1}$

4 0

3 years ago

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and

swat32

Answer:

$\%\ mass\ of\ CaCO_3=93.37\ \%$

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = 293.15 K

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.mmHg/K.mol

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K

⇒n of $CO_2$ produced = 0.0493 moles

According to the reaction:-

$CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2$

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of $CaCO_3$ = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0493\ mol= \frac{Mass}{100.0869\ g/mol}$

$Mass_{CaCO_3}=4.93\ g$

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

$\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100$

$\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100$

$\%\ mass\ of\ CaCO_3=93.37\ \%$

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3 years ago

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