All categories

3 years ago

If a bag of peas weighs 454 grams, how many peas would be in the bag?

Chemistry

1 answer:

OverLord2011 [107]3 years ago

6 0

In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

$454 g \times \frac{1pea}{0.1 g} = 4540pea$

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

$454 g \times \frac{1pea}{0.36 g} \approx 1261pea$

In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

You can learn more about conversion factors here: brainly.com/question/1844638

You might be interested in

In the name carbon dioxide what does the prefix of the second word indicate about a molecule of a dioxide A. It has two carbon a

brilliants [131]

Answer:

The answer is option B, that is, it has two oxygen atoms.

5 0

3 years ago

Cuál es la densidad (g/ml) de un liquido de batería de automóvil, si tiene un volumen de 0,125 L y una masa de 155g¿?

guajiro [1.7K]

Answer:

nnaaoooo sseeiiiiii!!!!!!!!!

7 0

3 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago

Which of the following liquids has the lowest viscosity?

zavuch27 [327]

Answer:

Ether and acetone are the liquids with the lowest viscosities at room temperature.

Explanation:

7 0

3 years ago

A 367.8 g sample of potassium chlorate was decomposed according to the following equation:

Pie

The answer is A i did this before hope this helps

7 0

3 years ago

​If a bag of peas weighs 454 grams, how many peas would be in the bag?

If a bag of peas weighs 454 grams, how many peas would be in the bag?