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3 years ago

I need help with this, please I need it ASAP

Chemistry

2 answers:

gizmo_the_mogwai [7]3 years ago

7 0

I’m pretty sure it’s B

AnnyKZ [126]3 years ago

3 0

I think it might be letter b. hope this helped have a wonderful day!!

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Which is the load D. A. B. C. E.

Paha777 [63]

Answer:

Explanation:

In the case of a seesaw, the object at the top is the effort and the object at the bottom is the load.

6 0

3 years ago

How many grams of a solution that is 5.2% sucrose by mass are needed to obtain 18g of sucrose

kirill115 [55]

Answer:

346.g of solution

Explanation:

In this case, if we have 5.2 % by mass it means that in 100 g of the solution we will have 5.2 g of glucose. Therefore we can do the calculation:

5.2 g of glucose = 100 g of solution

$18~g~of~glucose\frac{100~g~of~solution}{5.2~g~of~glucose}=346.15~g~of~solution$

So, if we need 8 g of glucose we had to have 346.15 g of solution

This logic can work for all types of solutions. By mass (as in this case), by volume or mass/volume.

I hope it helps!

3 0

4 years ago

A pharmacy intern is asked to prepare 3 L of a 30% w/v solution. T he pharmacy stocks the active ingredient in 8-ounce bottles o

MariettaO [177]

Answer: The number of bottles that will be needed are 6

Explanation:

We are given:

Amount of solution, the intern is asked to prepare = 3 L = 3000 mL (Conversion factor: 1 L = 1000 mL)

Strength of solution needed = 30 % (w/v)

This means that in 100 mL of solution, the solute present is 30 grams

So, in 3000 mL of solution, the solute present will be = $\frac{30}{100}\times 3000=900g$

Active ingredient present in 1 bottle = 8 ounce of 70 % (w/v)

Conversion factor used: 1 ounce = 29.57 mL

So, $8ounce\times \frac{29.57mL}{1ounce}=236.6mL$

Amount of active ingredient present in 1 bottle = $236.6\times \frac{70}{100}=165.6g$

To calculate the number of bottles, we need to divide the total amount of solution needed by the amount of active ingredient present in 1 bottle, we get:

$\text{Number of bottles}=\frac{\text{Amount of solution to prepare}}{\text{Amount of active ingredient in 1 bottle}}$

Putting values in above equation, we get:

$\text{Number of bottles}=\frac{900g}{165.6g}\\\\\text{Number of bottles}=5.43\approx 6$

Hence, the number of bottles that will be needed are 6

6 0

3 years ago

How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel

Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

6 0

3 years ago

!!!!!!!I REALLY NEED HELP PLEASE!!!!!!!!!!!

Brut [27]

Answer:

2:1 Hydrogen:Oxygen for all three answers.

Explanation:

Maltose has 22 hydrogen and 11 oxygen,

glucose and fructose both have 12 hydrogen and 6 oxygen, and water has 2 hydrogen and 1 oxygen.

3 0

3 years ago