All categories

3 years ago

In a 28 g serving of cheese curls there are 247mg of sodium. How much sodium is in a 12.5 ounce bag

Chemistry

1 answer:

Marina86 [1]3 years ago

6 0

Answer:

There should be about 3136.9mg of Na in a 12.5oz bag of cheese curls.when rounding if not rounded it would be 3126.04082143mg

Explanation:

12.5oz=354.4g

354.4/28=12.7

12.7*247=3136.9

You might be interested in

What is the density of a substance that has a mass of 20 g and a volume of<br> 10 mL?

Paha777 [63]

Answer:

That unknown substance is water

Explanation:

3 0

2 years ago

Why is salt water intrusion a big problem in florida?

Schach [20]

Its leading to contaminated drinking water.

7 0

2 years ago

Read 2 more answers

An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 39 mg every 30 min. The IV bag

Korvikt [17]

Answer:

The Flow rate = 0.0208 mL/min

Explanation:

Data provided:

Rate of dose = 39 mg every 30 min = (39/30) mg/min = 1.3 mg/min

also,

125mg of methylprednisolone is present in every 2 mL

thus,

concentration = (125/2) mg/mL = 62.5 mg/mL

Now,

The flow rate is given as:

Flow rate = Rate / Concentration

on substituting the respective values, we get

Flow rate = (1.3 mg/min) / (62.5mg/mL)

The Flow rate = 0.0208 mL/min

3 0

3 years ago

What does squeezing do to the particles of snow or ice?

3241004551 [841]

Answer:

crushes the ice to compaction and satasfation

3 0

3 years ago

Read 2 more answers

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0

3 years ago