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Ghella [55]
3 years ago
15

What does your immune system do? a. makes you sick b. keep your brain sharp c. protect you from illness d. make energy for your

body
PLEASE MARK AS BRAINLIEST
Chemistry
2 answers:
DerKrebs [107]3 years ago
7 0
C keeps you from I’ll was I agree
Dima020 [189]3 years ago
6 0

Answer:

C.KEEPS YOU FROM ILLNESS

Explanation:

THE WORD IMMUNE MEANS TO PROTECT THUS THE IMMUNE SYSTEM IS TO PROTECT YOUR BODY AGAINST PATHOGENS AND DEASISES CAUSING GERMS....

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kogti [31]
The physical process of evaporation involves water and the sun
7 0
3 years ago
Can someone pls tell me the answer to these 2 fill in the blanks I will surely mark him as the brain list answer
Elena-2011 [213]
Question 22: conclusion
Question 23: analyze
7 0
3 years ago
If ammonium phosphate reacts with sodium chloride in aqueous solution, what are the products?
Ilia_Sergeevich [38]

Answer:

(NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)

Explanation:

Hello there!

In this case, according to the given information, we can set up the appropriate chemical equation when ammonium phosphate reacts with sodium chloride in aqueous solution:

(NH_4)_3PO_4(aq)+NaCl(aq)\rightarrow NH_4Cl(aq)+Na_3PO_4(aq)

Which stands for a double replacement reaction, whereby ammonium changes phosphate to chloride and sodium changes chloride to phosphate on the products side. In addition, we can balance the aforementioned equation as shown below:

(NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)

Regards!

4 0
3 years ago
What is the percent of carbon-14 remaining after two half-lives?
Maslowich
After one half-life, one half is remaining.  After two half-lives one fourth is remaining.  So that would be 25%

7 0
4 years ago
For the following reaction, if 19.7 g of H2, is reacted with excess CO in the laboratory, and 144.5 g of CH3OH is produced, what
GuDViN [60]

Step 1

<em>The reaction involved:</em>

CO + 2 H2 → CH3OH (completed and balanced)

------------

Step 2

<em>Data provided:</em>

19.7 g H2 (the limiting reactant)

Excess reactant = CO

144.5 g CH3OH = actual yield

----

<em>Data needed:</em>

The molar masses of:

H2) 2.00 g/mol

CH3OH) 32.0 g/mol

-----------

Step 3

The theoretical yield:

By stoichiometry,

CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)

2 x 2.00 g H2 --------- 32.0 g CH3OH

19.7 g H2 --------- X

X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2

X = 157.6 g CH3OH (The theoretical yield)

-----------

Step 4

The % yield is defined as follows:

\begin{gathered} \text{ \%yield = }\frac{Actual\text{ yield}}{Theoretical\text{ yield}}x100\text{ } \\ \text{ \%yield = }\frac{144.5\text{ g}}{157.6\text{ g}}x100\text{ = 91.7 \% = 92 \% approx.} \end{gathered}

Answer: d. 93% (it is the nearest value in comparison to my result)

6 0
2 years ago
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