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Akimi4 [234]
3 years ago
11

This worm infects humans by getting into the blood stream. It gets a

Chemistry
1 answer:
liraira [26]3 years ago
5 0
Pretty sure it’s C- hope this helps❤️
You might be interested in
Please help! This is for the 9.06 Acid Neutralization Lab! Due today!
Volgvan

hi there bestie [i think its 3]

5 0
3 years ago
Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the
bekas [8.4K]
The correct values I believe would be a=1 b=-2 and c=-3.
8 0
3 years ago
The compound magnesium nitrate has the formula Mg(NO3)2.
Ludmilka [50]

Explanation:

Answer

Open in answr app

Correct option is

C

148

1 atom of Mg=1x24=24

2 atom of N=2x14=28

6 atom of O=6x16=96

Total mass=148

4 0
3 years ago
How many total ions are present in 347g of cacl2?
vladimir1956 [14]

In 1 mole of CaCl_{2}, there are 3 moles of ions, 1 mole of Ca^{2+} and 2 mole of Cl^{-1}.

CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-}

Molar mass of CaCl_{2} is 110.98 g/mol. Calculating number of moles from given mass as follows:

n=\frac{m}{M}=\frac{347 g}{110.98 g/mol}=3.12 mol

Thus, number of moles of ions will be 3\times 3.12 mol=9.38 mol.

Since, 1 mole of any substance has 6.023\times 10^{23} units of that substance where 6.023\times 10^{23}  is Avogadro's number.

Thus, 9.38 mol of ions will have 9.38\times 6.023\times 10^{23}=5.65\times 10^{24} number of ions.

Therefore, total number of ions in 347 g of CaCl_{2} is  5.65\times 10^{24}.


8 0
3 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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